The field Q(2cos(pi/n)), its Galois group and length ratios in the regular n-gon
aa r X i v : . [ m a t h . G R ] M a r KarlsruheOctober 03 2012Revised: March 03 2017
The field Q ( (cid:0) π n (cid:1) ) , its Galois group, and length ratios in theregular n-gon Wolfdieter L a n g Abstract
The normal field extension Q ( ρ ( n )), with the algebraic number ρ ( n ) := 2 cos (cid:16) πn (cid:17) , for n ∈ N ,is related to ratios of the lengths between diagonals and the side of a regular n − gon. This has beenconsidered in a paper by P. Steinbach. These ratios, numbered k = 1 , ..., n −
1, are given by
Chebyshev polynomials S ( k − , x = ρ ( n )). The product formula for these ratios was found by Steinbach and isre-derived here from a known formula for the product of
Chebyshev S − polynomials. It is shown thatit follows also from the S − polynomial recurrence and certain rules following from the trigonometricnature of the argument x = ρ ( n ). The minimal integer polynomial C ( n, x ) for ρ ( n ) is presented, andits simple zeros are expressed in the power-basis of Q ( ρ ( n )). Also the positive zeros of the Chebyshev polynomial S ( k − , ρ ( n )) are rewritten in this basis. The number of positive and negative zeros of C ( n, x ) is determined. The coefficient C ( n,
0) is computed for special classes of n values. Theoremson C ( n, x ) in terms of monic integer Chebyshev polynomials of the first kind (called here ˆ t ) are given.These polynomials can be factorized in terms of the minimal C -polynomials. A conjecture on thediscriminant of these polynomials is made. The Galois group is either Z δ ( n ) , the cyclic group of theorder given by the degree δ ( n ) of C ( n, x ), or it is a direct product of certain cyclic groups. In orderto determine the cycle structure a novel modular multiplication, called M odd n is introduced. On thereduced odd residue system M odd n this furnishes a group which is isomorphic to this Galois group.
Length ratios between diagonals and the side of a regular n − gon (called diagonal/side ratios, abbreviatedDSRs) have been considered by Steinbach [29]. He gave a product formula for these ratios (called byhim diagonal product formula (DPF)), and for the pentagon and heptagon details were given. In thepentagon case the quadratic number field Q ( √
5) with the basis < , ϕ > for integers of this field turnsup. Here ϕ , the golden section, is identified with ρ (5) which is the length ratio between any of the twodiagonals and the pentagon side. For the heptagon there are two different diagonal length and the tworatios between them and the side length have been called ρ := ρ (7) and σ . The DPF allowed to reduceall products and quotients of ρ and σ to Q -vectors with the basis < , ρ, σ > . For example, ρ = 1 + σ ,and one can use instead the power basis < , ρ, ρ > of the algebraic number field Q ( ρ (7)) (the use of thesame acute bracket notation for different bases should not lead to a confusion). The minimal polynomialof ρ (7), in the present work called C(7,x), is x − x − x + 1, and this was also given in [29]. Therefore,[ Q ( ρ (7)) : Q ] = 3, which is the degree of this field extension. For odd n , n = 2 k + 1, Steinbach gavea polynomial with one of its roots ρ (2 k + 1). In the present work the minimal polynomial for ρ ( n ),called C(n,x), for every n ∈ N , is given in terms of the known one for cos (cid:18) πN (cid:19) , called Ψ( N, x ), with N = 2 n . The connection between Ψ( N, x ) and divisor product representations of N , and to Chebyshev T − polynomials has been worked out earlier by the author. See the links under OEIS [27] A181875 andA007955 (in the sequel OEIS Anumbers will appear without repeating this reference). C ( n, x ) turns out [email protected] ,
1o be an integer polynomials of degree δ ( n ) = A055034( n ). All its zeros are known, and they are simple. Q ( ρ ( n )) is the splitting field for the minimal polynomial C ( n, x ) of ρ ( n ). It is a normal extension ofthe rational field. These C -polynomial zeros are written in the power basis of Q ( ρ ( n )) with the help ofcertain scaled Cheyshev T − polynomials (called here ˆ t with their integer coefficient array A128672). Thisarray can also be used to rewrite the positive zeros of the Chebyshev S ( n − , x ) polynomial, related tothe DSRs, in this power basis. The use of S − polynomial technology allows also for a re-derivation of theproduct formulae (DPF) for the length ratios between the regular n − gon diagonals and side.In section 2 S ( k − , ρ ( n )) is shown to yield the DSRs of the regular n -gon. The DPF is also re-derivedthere, and the independent products are extracted. This is called reduced algebra over Q . In section 3the minimal polynomial C ( n, x ) is presented, and its zeros are related to the power basis of the algebraicnumber field Q ( ρ ( n )). In section 4 the Q -automorphisms of this field, the Galois group G ( Q ( ρ ( n )) / Q ),is treated. It is the cyclic group Z δ ( n ) , except for infinitely many n -values where it is the direct productof cyclic groups, hence non-cyclic. For n from 1 ..
100 there are 30 non-cyclic groups. A novel modularmultiplication on the odd numbers, called M odd n , is introduced which serves to determine the cyclegraph structure of this Galois group. The multiplicative group M odd n is isomorphic to this Galois group.On the reduced odd residue system M odd n this multiplicative group is isomorphic to the Galois group.
Motivated by a paper of Steinbach [29] we consider the diagonals and the side in the regular n − gon(inscribed in the unit circle). The vertices on the unit circle are called V ( n ) k , k = 0 , ...., n − , n = 2 , , ... ; V ( n )0 has coordinates (1 , s ( n ) = 2 sin (cid:16) πn (cid:17) . Thelength of V ( n )0 V ( n ) k is d ( n ) k = 2 sin (cid:18) π kn (cid:19) , for k = 1 , ..., n −
1, hence d = s ( n ). We only need one sideand the diagonals in the upper half plane, i.e. it suffices to consider k ∈ { , ..., ⌊ n/ ⌋} . For even n , thelargest diagonal ( k = n Fig. 1 , and
Fig. 2 , for the case n = 10 (decagon), and for n = 11 (hendecagon), respectively. The length ratios of interest, the DSRs,are R ( n ) k = d ( n ) k d ( n )1 for the given k − values. Here Chebyshev S − polynomials in their trigonometric formenter the stage (see A049310 for their coefficients): R ( n ) k = S ( k − , ρ ( n )) , (1)where we use the second ratio (for the smallest diagonal) ρ ( n ) := R ( n )2 = 2 cos (cid:16) πn (cid:17) . Remember that S ( n, x ) := U ( n, x/
2) with
Cheyshev U − polynomial (second kind). It is well known that the zeros of thepolynomials S ( n − , x ) are x ( n − k, ± = ± (cid:18) k πn (cid:19) , for k = 1 , ..., (cid:6) n − (cid:7) . Note that in the even n casethe zero x = 0 appears twice from ±
0. The S − polynomials are orthogonal polynomials defined on thereal interval [ − , +2]. Hence their zeros are guaranteed to be simple. They belong to the Jacobi class.Due to (anti)symmetry it is sufficient to consider only the positive zeros (we disregard x = 0 in the even n case). Now also Chebyshev T − polynomials (first kind; see the coefficient table A053120) enter, whichare written in terms of S − polynomials. (The T − polynomials appear as trace polynomials in a 2 × x ( n − k ≡ x ( n − k, + = 2 T ( k, ρ ( n ) /
2) =: ˆ t k ( ρ ( n )) ≡ ˆ t ( k, ρ ( n )) (2)= S ( k, ρ ( n )) − S ( k − , ρ ( n )) , k = 1 , ..., (cid:22) n − (cid:23) . (3)2ote that these positive zeros decrease with k , and that ˆ t k is an integer polynomial of degree k because the S − polynomials are integer. The coefficient array for these monic ˆ t -polynomials is A127672. In [1] thesepolynomials are called Chebyshev C -polynomials, but we use the letter C for the minimal polynomialsof ρ ( n ). The zeros can be replaced by DSRs from the above given eq. (1). Positive zeros of S ( n − , x ) from DSRs: x ( n − k = R ( n ) k +1 − R ( n ) k − . (4)Figure 1: Decagon n = 10 Figure 2: Hendecagon (n=11)As an aside we give the factorization of S ( n − , x ) in terms of the polynomial P ( k, x ) := k Y l =1 (cid:18) x − (cid:18) l π k + 1 (cid:19)(cid:19) with positive zeros. S ( n − , x ) = x Θ( n − P (cid:18)(cid:22) n − (cid:23) , x (cid:19) ( − ⌊ n − ⌋ P (cid:18)(cid:22) n − (cid:23) , − x (cid:19) , (5)where θ ( n −
1) := 0 if n is odd, and 1 if n is even. This factorization is also considered in [4] for the U ( n, x ) polynomials for even and odd n separately. For example, for the heptagon n = 7 one finds thefactorization of S (6 , x ) with the following real polynomial P (3 , x ). P (3 , x ) = x − (2 σ (7) − x + 2 ρ (7) x − , (6)with ρ (7) = R (7)2 and σ (7) := R (7)3 (see also the introduction). The three positive zeros are, written interms of the DSRs, ρ (7) , σ (7) − σ (7) − ρ (7). Here we used the general translation formulae eq.(4) to rewrite the positive zeros of S ( n − , x ) in terms of DSRs. One should also write this in terms ofpowers of ρ (7) with the help of eq. (1) (see eq. (7) below).Each P (cid:18)(cid:22) n − (cid:23) , x (cid:19) can be considered as characteristic polynomial for a ( (cid:4) n − (cid:5) + 1)-term recurrencesequence { f k } ∞ k =0 in the integral domain of Q ( ρ ( n )) (represented by a δ ( n )-tuple of ordinary (rational)integers ( A ,k , A ,k , ..., A δ ( n ) ,k ). Here δ ( n ) is the degree of the minimal polynomial of ρ ( n ) which will bediscussed in sect. 3. In fact, one has (cid:4) n − (cid:5) such δ ( n )-tuple sequences corresponding to the independentinputs. As a simple example take the n = 5 pair ( δ (5) = 2) of sequences corresponding to P (2 , x ) = x − (2 ϕ − x + 1, with the golden section ϕ := ρ (5) and the simplest input. This is S ( k, √
5) = A ,k A ,k ϕ with input S ( − , √
5) = 0 , S (0 , √
5) = 1. See A005013( k + 1) ( − k , and 2 A147600( k − A ,k and A ,k , respectively. 3or later purposes it is also useful to give a dictionary between the positive zeros of S ( n − , x ), i.e. thoseof P ( (cid:4) n − (cid:5) , x ), and powers of ρ ( n ). One direction follows from the above given formula, eq. (2), for x ( n − k in terms of the monic integer polynomials ˆ t ( k, ρ ) = 2 T ( k, ρ/ n has herebeen omitted. Positive zeros of S ( n − , x ) from ρ ( n ) -powers: x ( n − k = ⌊ k ⌋ X l =0 ( − l (cid:18) k − ll (cid:19) ρ ( n ) k − l . (7)The coefficients are given in the triangle A127672. E.g. , x = 5 ρ − ρ + ρ (omitting n ). Forthe heptagon this shows that the three positive zeros of S (6 , x ) are ρ (7) , σ (7) − ρ (7) − σ (7) − ρ (7) = ρ (7) − ρ (7) − Q -vector spacebasis. Note that the n dependence is via the ˆ t -polynomial variable ρ ( n ) (which satisfies C ( n, ρ ( n )) = 0 bydefinition of the minimal polynomial C ). For example, it is true for all n that the second largest positivezero of S ( n − , x ), namely x ( n − , is always ρ ( n ) −
2. Hence, from above, R ( n )3 − ρ ( n ) −
2; orif one calls R ( n )3 =: σ ( n ) then ρ ( n ) = σ ( n ) + 1. This has been noted for n = 7 above but is holdsin general, showing that σ ( n ) is algebraically dependent on ρ ( n ). This fact has been observed already in[29] where it appeared as a special product formula (see the later DPF eq. (13), m = k = 2, or eq. (11)with k = 2).The inverse formulae (DSRs in terms of ρ -powers) have already been given above in eq. (1). Here oneuses the S -triangle A049310 for the translation. E.g. , R ( n )4 = − ρ ( n ) + ρ ( n ) = ρ ( n ) ( ρ ( n ) − ρ ( n ) ( σ ( n ) −
1) . Of course, R ( n )4 is interesting only for n ≥ S ( n − , x ), i.e. , (cid:4) n − (cid:5) , is less or equal to (cid:4) n (cid:5) , the numberof DSRs for diagonals in the upper half plane including the negative real axis. The zero x = 1 neverappears. As can be seen in the n = 7 case these zeros can nevertheless be used as Q -vector space basis (insect. 3 it will become clear that the degree δ (7) is also 3). In the case n = 9 the two numbers are both 4,but the degree δ (9) is 3, hence only three of the zeros and three of the DSRs are rationally independent.To wit: x (8) = − x (8)3 + 3 x (8)1 − x (8)2 and R (9)4 = R (9)1 + R (9)2 = 1 + ρ (9).Because of symmetry only DSRs for diagonals of the upper half plane (including the negative real axis), i.e. , k ∈ { , , ..., j n k } , are of interest. The reduction for other k values, also negative ones, isaccomplished by the rules o) R ( n ) n + k = − R ( n ) k , i) R ( n ) −| k | = − R ( n ) | k | , ii) for k ∈ { j n k + 1 , ..., n } : R ( n ) k = R ( n ) n − k . (8)These rules follow from eq. (1) with the trigonometric definition (with the specific value of ρ ( n )) of the S -polynomials, and therefore also negative values for the DSRs show up (the interesting DSRs are, ofcourse, positive). i ) follows also from S ( −| n | , x ) = − S ( | n | − , x ) which derives from a backward use ofthe recurrence (given later in eq. (12)). This rule does therefore not depend on the special choice of thevariable x , in contrast to the rules o ) and ii ). With negative k one counts the diagonals in the clockwisedirection. ii) is used to translate from the lower to the upper half-plane.The product formula for Chebyshev S -polynomials of different degree, but with the same argument, iswell known. See e.g. , [1], p.782, 22.7.25, for the U -polynomials, and replace the T -polynomials by the S -polynomials via the trace formula given in eqs (2) and (3). S ( m − , x ) S ( n − , x ) = 14 (( x ) −
1) [ S ( n + m, x ) − S ( n + m − , x ) − S ( n − m, x ) + S ( n − m − , x ) ] n ≥ m . (9)4or our purpose x = ρ ( n ) = R ( n )2 = ± n with m if the rules for negative indices on the S -polynomials, statedabove, are employed. Here we restrict to n ≥ m in order to have non-negative indices. Thus the productformula for the DSRs, now with x = ρ ( n ), but without using its specific value, is(4 − ( R ( n )2 ) ) R ( n ) m R ( n ) k = R ( n ) k − m +1 − R ( n ) k − m − − R ( n ) k + m +1 + R ( n ) k + m − , k ≥ m > . (10)For m = 0 this becomes trivial. This is not yet the DPF given by Steinbach in [29] which linearizesthe product of R ( n ) m R ( n ) k . In order to eliminate the pre-factor (4 − ρ ( n ) ) one can use the three termrecurrence relation of the orthogonal S -polynomials, written for the DSRs by eq. (1) as a special productformula [ Rec , k ] ( n ) : R ( n )2 R ( n ) k = R ( n ) k +1 + R ( n ) k − , k ≥ . (11)From now on we also use the specific form of ρ ( n ). Thus also rules o ) and ii ) of eq. (8) will beapplicable. For k = 0 one uses (see ii) ) R ( n ) − = − R ( n )1 = −
1, and it becomes trivial. For k = 2it shows that ρ ( n ) = σ ( n ) + 1 for all n , if one uses σ ( n ) := R ( n )3 . This recurrence can now beused twice as R ( n )2 ( R ( n )2 R ( n ) k ) in eq. (10) to produce the following recurrence for two step differences of R ( n ) m R ( n ) k =: p ( n,m ) k ≡ p k ( p k +2 − p k ) − ( p k − p k − ) = c k − c k − , (12)where we used the abbreviation c k ≡ c ( n,m ) k := R ( n ) k + m +1 − R ( n ) k − m +1 . This shows that p k +2 − p k − c k is k independent. From the inputs p − p − c = 0, due to the recurrence relation eq. (11), and p − p − − c − = 0 with the help of the rule i) from eq. (8), this leads to the recurrence p k +2 − p k = c k ,with the inputs p = 0 and p = − p = R ( n ) m . The solution for p k is found for even and odd k separately,where again the rules for negative indices are employed. Both solutions can then be combined as DPF [ m , k ] ( n ) : R ( n ) m R ( n ) k = k − X j =0 R ( n ) m + k − (2 j +1) , ≤ m ≤ k . (13)This is finally the DPF found by Steinbach in [29] and gives a linearization of the DSR products. Weneed only to consider m ≥ m = 1 becomes trivial), and due to the symmetry of this formula underthe transformation k → n − k (using the rules of eq. (8)) is suffices to consider k ∈ { , , ... (cid:4) n (cid:5) } . Henceone has only to consider the (cid:18)(cid:4) n (cid:5) (cid:19) products for 2 ≤ m ≤ k ≤ (cid:4) n (cid:5) .The DPF formula looks un-symmetric with respect to m ↔ k , but it is, in fact, symmetric because m − X j =0 R ( n ) k + m − (2 j +1) for k < m reduces, due to cancellations after using the rule i ) from eq (8), to theexpected sum with only k terms. One can see this for even and odd m > k separately, rememberingthat R ( n )0 = 0 in the former case.The idea is to work out, for a given n ≥
4, all DPFs of interest, using the rules from eq. (8), especially i ) and ii ) in order to write all (cid:18)(cid:4) n (cid:5) (cid:19) products as linear combinations of the DSRs. Example 1 : n = 7 heptagon , treated in detail in reference [29]. Here we recapitulate and link to OEIS[27] sequences (the analoga of Fibonacci numbers in the pentagon case). Superscripts and arguments 7are suppressed. [2 , ρ = σ + 1. [2 , ρ σ = σ + ρ + 0. [3 , σ = ρ + σ + 1. Here, R = R was used. This shows that the Q -vector space basis is at most < , ρ, σ > . It will be shown to be indeedthe heptagon basis in sect. 4, where the power basis will be used instead.With these DPFs one can compute all powers of interest in the heptagon basis < , ρ, σ > (this hasalready been done explicitely for σ in [29], p. 28). 5 k = A052547( k −
2) 1 + A052547( k − ρ + A006053( k ) σ , k ≥ ρ − k = A077998( k ) 1 + A077998( k − ρ − A006054( k + 1) σ , k ≥ σ k = A106803( k −
1) 1 + A006054( k − ρ + A106803( k ) σ , k ≥ σ − k = ( σ − ρ ) k = A052547( k ) 1 − A006053( k + 1) ρ − A052547( k − σ , k ≥ ρ σ ) k = ( ρ + σ ) k = A120757( k ) 1 + —A006054( k − | ρ + 4 A181879( k ) σ , k ≥ ρ σ ) − k = ( ρ + σ ) − k = A085810( k ) 1 + ( − k A181880( k − ρ + ( − k +1 A116423( k + 1) σ , k ≥ a b ρ + c ρ = A B ρ + σ and find with N ( a, b, c ) = a − b − c − a b − a b c + a b + b c + 2 a c − a c + 2 b c A = 1 N ( a, b, c ) ( a − b + a b + 2 a c − b c ) , B = 1 N ( a, b, c ) ( b − c + a b ) , (14) C = 1 N ( a, b, c ) ( c − b + a c − b c ) . (15) Example 2: n = 9 ( Enneagon ), where we use ρ = R (9)2 , σ = R (9)3 , τ = R (9)4 . [2 ,
2] : ρ = 1 + σ ,[2 ,
3] : ρσ = ρ + τ , [2 ,
4] : ρτ = σ + τ , [3 ,
3] : σ = 1 + τ + σ , [3 ,
4] : σ τ = ρ + σ + τ ,[4 ,
4] : τ = 1 + ρ + σ + τ . This DSR-algebra (over Q ) shows that not all ratios (including R ( n )1 = 1)are linear independent: τ ( ρ −
1) = σ = ρ − ρ + 1) ( ρ − i.e. , τ = ρ + 1. Therefore, ρ = ρ ( σ + 1) = 2 ρ + τ + 2 = 3 ρ + 1 . In sect. 3 we will see that the algebraic number ρ (9)has degree 3, and its minimal polynomial is indeed C (9 , x ) = x − x −
1. The enneagon basis isthus < , ρ, σ > which can be related to the power basis < , ρ, ρ > (remember that we use the samenotation for different bases). The reduced DSR-algebra (the algebra modulo C (9 , ρ ) = 0 ) is ρ = 1 + σ , ρ σ = 1 + 2 ρ , and σ = 2 + ρ + σ . Remark 1: [2 ,
2] in eq. (13) becomes ( R ( n )2 ) = R ( n )3 + 1, showing that for all interesting values n ≥ σ ( n ) := R ( n )3 = ρ ( n ) − k = 3, and it will be used insect. 3 as R ( n )3 rewritten in terms of the power basis of Q ( ρ ( n )). Remark 2:
For n = 4 and n = 5 the second diagonal (defining R ( n )3 ) is not of interest because it isin the lower half plane. According to eq. (8), ii) R (5)2 = R (5)3 , hence σ (5) = ρ (5) and the generalrelation [2 ,
2] from eq. (13) between σ ( n ) and ρ ( n ) leads to ρ (5) = ρ (5) −
1, the golden section formula ρ (5) = ϕ := (1 + √ /
2. In sect. 2 it will be seen that the minimal polynomial for ρ (5) is indeed C (5 , x ) = x − x − S -polynomials and the rules of eq. (8) onecan prove it directly by induction from the S − recurrence including negative indices in accordance withthese rules. Proposition 1:
The DPFs [ m, k ] ( n ) , eq. (13) but now for all 2 ≤ m, ≤ k , follow from the recurrence, eq. (11), andthe rules i) and ii) from eq. (8). Proof:
This is shown by double induction. First one shows this for given m ≥ k ≥ k . Then by induction over m for all k ≥ k -induction uses as starter [2 , ( n ) which is the recurrence. Assume that [2 , k ′ ] ( n ) is true for all k ′ = 2 , , ..., k −
1. Now (we omit the superscripts) R R k = R ( R R k − − R k − ) from the recurrence.Then use for both terms the induction hypothesis. Note that one obtains for the first term k −
1, andfor the second one k − ii ) only two terms survive in each case. (The sametype of cancellation was at work when we remarked above on the symmetry of the DPF formula in m and k .) Therefore, one obtains R ( R k − R k − + 0) − ( R k − − R k − + 0) which becomes, after use ofthe recurrence applied twice, R k +1 + R k − . This is indeed the desired result for [2 , k ] if one uses againrule ii) to get a truncation of the sum after two terms.6he m -induction uses as starter [2 , k ] for all k ≥
2, which has just been established. Then assume that[ m ′ , k ] is true for all m ′ = 2 , , ..., m − R m R k = R R m − R k − R m − R k from the recurrence.Assume the induction hypothesis for each term, obtaining, after use of the recurrence for the first sum, k − X j =0 ( R m + k − (2 j +1) + R m + k − − (2 j +1) ) − k − X j =0 ( R m − k − (2 j +1) which is indeed the assertion after cancellationof the last two sums. (cid:3) As mentioned in the example 2 the DSR-algebra turns sometimes out to be reducible because some DSRscan be expressed as rational linear combinations of other ones. Later it will become clear that thishappens precisely whenever (cid:4) n (cid:5) − δ ( n ) >
0, and this is the number of linear dependent DSRs. See
Table 1 for details for n = 3 , ..., sects. 3 and that it is simpler to use the power basis of Q ( ρ ( n )) and the minimal C ( n, x )-polynomials of the algebraic number ρ ( n ) instead of the DSR-algebra. Then one obtains automaticallythe reduced algebra. ρ ( n ) , its zeros, absolute term and factoriza-tion The minimal polynomial of an algebraic number α of degree d α is the monic, minimal degreerational polynomial which has as root, or as one of its roots, α . This degree d α is 1 iff α is rational, andthe minimal polynomial in this case is p ( x ) = x − α . For the notion ‘minimal polynomial of an algebraicnumber’ see, e.g. , [22], p. 28 or [25] p. 13.For the algebraic number ρ ( n ) := 2 cos (cid:16) πn (cid:17) , for n ∈ N , the degree is δ (1) = 1, and δ ( n ) = ϕ (2 n )2for n ≥ Euler ’s totient function ϕ ( n ) = A n ). This is the sequence A055034( n ). Thesequence of allowed δ values is given in A207333. The array with the indices of the polynomials for givenallowed δ values is shown in A207334. Of course, δ is not multiplicative, e.g. , 2 = δ (6) = δ (3) δ (3) =1 · (cid:18) πn (cid:19) which are found, e.g. , underA181875/ A181876, and they have been called there Ψ( n, x ). See also [17], and [22], Theorem 3 .
9, p. 37,for the degree d ( n ) of these polynomials. From the trivial identity cos (cid:16) πn (cid:17) = cos (cid:18) π n (cid:19) one finds theminimal polynomial of 2 cos (cid:16) πn (cid:17) , called here C ( n, x ), from C ( n, x ) = 2 δ ( n ) Ψ (cid:16) n, x (cid:17) . (16)Therefore the above formula for δ ( n ) derives from d (2 n ). These polynomials are given for n = 1 , ..., Table 2 , and the first 15 rows of their coefficient array are shown in
Table 3 . Concerning the parity ofthe degree δ one has the following lemma. Lemma 1: Parity of the degree δ δ ( n ) is odd iff n = 1 , , and p e , with an odd prime p of the form 4 k + 3, k ∈ N , and e ≥ Proof:
The case n = 1 is clear by definition. For n = 2 one uses the prime number factorization n =2 β Q Nj =1 p e j j , with β ≥ e j = 1. It follows from the definition that δ ( n ) = 2 β − N Y j =1 ( p j − p e j − j .For β ≥ β = 0 one needs N = 1, otherwise this will be even. For N = 1 oneneeds p −
12 to be odd, i.e. , p = 4 k + 3 with k ∈ N . If β = 1 one needs N = 0 in order to have anodd value. Thus n = 2. (cid:3)
7n order to relate to sect. 2 we are interested here in n ≥
4. It turns out that these polynomials are infact integer (not only rational) polynomials. The proof will use the following lemma based on our divisorproduct representation paper [14].
Lemma 2: Minimal polynomial C ( n , x ) written as a rational function. C ( n, x ) = p ( d p ( n ) , x ) q ( d q ( n ) , x ) , (17)with monic integer polynomials p and q with a certain degree d p ( n ) and d q ( n ) = δ p ( n ) − δ ( n ), respec-tively. Proof : Ψ (cid:0) n, x (cid:1) is obtained from the unique divisor product representation dpr (2 n ) defined in [14]by replacing each a ( k ) in the numerator, as well as in the denominator, by t ( k, x ) which is given as adifference of monic integer polynomials ˆ t which have been defined already in eq. (2), multiplied with acertain prefactor. t ( k, x k (cid:0) ˆ t ( k + 1 , x ) − ˆ t ( k − , x ) (cid:1) if k is even , k −
12 +1 (cid:0) ˆ t ( k +12 , x ) − ˆ t ( k − , x ) (cid:1) if k is odd . (18)The monic Ψ(2 n, x ) polynomials have degree d (2 n ) (see e.g. , [13]) which implies that all the prefactorsin this replacement of the a ( k )s in dpr (2 n ) have to become 1 / d (2 n ) . (This could be formulated as aseparate lemma). E.g. , n = 34 with dpr (34) = a (34) a (1) a (17) a (2) has from the numerator the factor 1 / and from the denominator 2 , fitting with 1 / d (34) because d (34) = 8. Therefore, one may in thecalculation of C ( n, x ), found above from Ψ(2 n, x ), forget about these prefactors in the replacementsaltogether. Thus the numerator, resp. denominator, is a product of monic integer polynomials ˆ t leadingto the monic integer polynomials p , resp. q , of a certain degree d p ( n ), resp. d q ( n ). (One could givemore details on these degrees but this is not important here. Trivially, d p ( n ) − d q ( n ) = d (2 n ). Forthe given example n = 34: (18 + 1) − (9 + 2) = 8 from the degrees of the ˆ t polynomials.) BecauseC(n,x) is a minimal polynomial this rational function allows polynomial division without remainder. (cid:3) Proposition 2: C ∈ Z [ x ] C ( n, x ), the minimal polynomial of ρ ( n ) = 2 cos (cid:16) πn (cid:17) , is an integer monic polynomial. Proof:
From lemma 2 we have q ( d q ( n ) , x ) C ( n, x ) = p ( d p ( n ) , x ) which leads by induction to the re-sult that the integer coefficients of the monic polynomials p and q imply integer coefficients for themonic polynomial C . Call these monic polynomials q ( N, x ) := P Nl =0 q l x l , C ( M, x ) := P Mk =0 c k x k and p ( N + M, x ) := P N + Ml =0 p l x l with q N = 1, c M = 1 and p N + M = 1. Collecting terms for x N + M − j , for j =(0) , , , ..., M , one obtains the formula for the C -coefficient c M − j in terms of lower indexed ones: c M − j = p N + M − j − j X k =1 c M − j − k q N − k . With this the inductive proof on j , using the integer coefficients of q, p andthe integer higher C -coefficients due to the induction hypothesis becomes obvious. The starting point isthe trivial j = 0 case. (cid:3) Next we give all the zeros of C ( n, x ). This follows directly from the knowledge of all zeros of Ψ(2 n, x )(see, e.g. , [13], from p. 473 of [30]) Proposition 3: All zeros of C C ( n, x ) = n − Y k =1 gcd ( k, n )=1 (cid:18) x − (cid:18) π kn (cid:19)(cid:19) , n ∈ N . (19)8n the product the index k starts with 1 because gcd (0 , n ) = 2 n , and it stops at n − n ≥ gcd ( n, n ) = n . One has to omit the even k values > n = 1 one takesonly l = 0 in the following product) C ( n, x ) = ⌊ n − ⌋ Y l =0 gcd (2 l +1 ,n )=1 (cid:18) x − (cid:18) π (2 l + 1) n (cid:19)(cid:19) . (20) Proof:
This is a simple consequence of the mentioned known results for Ψ(2 n, x ). It implies, en passant ,a formula for the known degree δ ( n ) (see A055034) in terms of the number of factors of this product. (cid:3) The following proposition lists the nonnegative and negative zeros of C for prime n . A vanishing zero(which will be counted later as positive) appears only for n = 2. Proposition 4: Non-negative and negative zeros of C ( p , x ) i) If n = 2 then C (2 ,
0) = 0 , and this is the only case with a vanishing zero. ii) If n is an odd prime 1 ( mod
4) then the number of positive and negative zeros coincides, and thisnumber is n − . These zeros are ( − k +1 (cid:0) π kn (cid:1) , with k values 1 , , ..., n − = δ ( n ). iii) If n is an odd prime 3 ( mod
4) then the number of positive zeros exceeds the number of negativeones by one, and the number of negative ones is n − . These zeros are ( − k +1 (cid:0) π kn (cid:1) with k values1 , , ..., n − , and an extra positive zero appears for k = n − = δ ( n ) . Proof:i)
Vanishing zeros require n ≡ mod
4) from kn = and k odd in eq. (20). A vanishing zero appearsfor n = 2. For other such n values, n = 4 K + 2, K ≥
1, the odd k = n > n , hence it doesnot appear in the product of eq. (20).For odd primes n = p = 2 q + 1 the product in eq. (20) is unrestricted, and there are ( q −
1) + 1 = p − δ ( p ) = ϕ (2 p )2 . The cos function will produce negative zeros for2 l + 1 p >
12 , i.e. , for 2 l ≥ q . ii) n = p = 4 K + 1 means that l = 0 , , ..., K − l = K, K + 1 , ..., K − K = n −
14 such zeros. For the l values leading to negativezeros one can use the formula cos (cid:18) πp (2 l + 1) (cid:19) = − cos (cid:18) πp ( p − (2 l + 1)) (cid:19) . In this way even multiplesof πp appear, and one produces, counted backwards, beginning with the largest l value, the new values2 , , ..., K . This proves ii) . As a test consider n = 13, K = 3: the 2 l + 1 values for the positivezeros are 1 , ,
5, and the ones for the negative ones are 7 , ,
11. The latter become (we use underliningto indicate that they come with a minus sign in the final formula) 6 , ,
2. Rearranged these values are1 , , , , ,
6, as given in ii) . iii) For n = p = 4 K + 3 a similar analysis leads to l = 0 , , ..., K , and l = K + 1 , ..., K for positive,and negative zeros, respectively. This shows that the number of negative zeros is K = p −
34 , and thenumber of positive ones exceeds the one for negative zeros by one. Again, the odd values leading to nega-tive zeros are transformed to even ones 2 , , ..., K with a minus sign in front of 2 cos. Take, e.g. , n = 19, K = 4, with the new k values of iii) ,
2, 3, 4 , ,
6, 7, 8 , (cid:3) For general values n one can also find the number of positive and negative zeros of C ( n, x ). In thenon-prime or non-power-of-2 case the gcd restriction in the product excludes certain l values. Thereforeone has to eliminate from the unrestricted 2 l + 1 values in the product of eq. (20) all odd multiples9or each odd prime dividing n . Multiples of 2 are irrelevant for even n , therefore only the odd primesdividing n are of interest. This shows that the prime factors of the squarefree kernel of n , denoted here by sqf k ( n ), i.e. , the largest squarefree number dividing n (see A007947) will be of interest. This squarefreekernel is also known as radical of n , denoted by rad ( n ). We will denote the set of primes of this kernelby sqf kset ( n ), E.g. , sqf kset (2 · ·
11) = { , , } , sqf k (2 · ·
11) = 66 (strip off all exponents inthe prime number factorization of n ). Because one may encounter multiple counting ( e.g. , 15 is hit bythe multiples of 3 as well as 5 for any n which has in its squarefree kernel set the primes 3 and 5, andwhich is larger than 16) one can employ PIE , the principle of inclusion-exclusion, e.g. , [5], p. 134, to geta correct counting.
Proposition 5: Number of positive and negative zeros of C ( n , x ) i) If n = 2 m , m ≥
1, one has for m = 1 a vanishing zero (see proposition 4 i) ). For m ≥ n δ (2 m ) = 2 m − = n ± (cid:16) πn (2 l + 1) (cid:17) , l = 0 , , ..., n − ii) If n = 1 then there is only the negative zero −
2. If n ≥ δ + ( n ), is δ + ( n ) = K ( n ) + M ( n ) X r =1 ( − r X (cid:22) (cid:18) L ( n ) p i · · · p i r − (cid:19)(cid:23) , (21)where M ( n ) (sometimes called ω ( n )) is in the odd n case the number of elements (cardinality) of theset sqf kset ( n ), denoted by | sqf kset ( n ) | , (see A001221). M ( n ) is for even n the number of odd primesin sqf kset ( n ), i.e. , | sqf kset ( n ) | −
1. This is because multiples of 2 are irrelevant here. The sum < i , i , ..., i r > extends over the (cid:18) Mr (cid:19) combinations 1 ≤ i < i < ... < i r ≤ M . Only the oddprimes from the set sqf kset ( n ) enter. The values K ( n ) and L ( n ) depend on the parity of n and they aregiven by α ) n even : K ( n ) = (cid:22) n − (cid:23) , L ( n ) = 2 K ( n ) + 1 , (22) β n odd , mod
4) : K ( n ) = n − , L ( n ) = 2 K ( n ) + 1 = n − , (23) β n odd , mod
4) : K ( n ) = n − , L ( n ) = 2 K ( n ) + 1 = n − . (24)Note that also negative values for the floor function may appear. In this way the pure prime case from proposition iii) If n ≥ δ − ( n ), is δ − ( n ) = N ( n ) + M ( n ) X r =1 ( − r X (cid:26)(cid:22) (cid:18) P ( n ) p i · · · p i r − (cid:19)(cid:23) − (cid:22) (cid:18) L ( n ) p i · · · p i r − (cid:19)(cid:23)(cid:27) , (25)where M ( n ) and L ( n ) are as above, and α ) n even : N ( n ) = n − − (cid:22) n − (cid:23) , P ( n ) = n − , (26) β n odd , mod
4) : N ( n ) = n − , P ( n ) = n − , (27) β n odd , mod
4) : N ( n ) = K ( n ) = n − , P ( n ) = n − . (28)10 roof:i) All l = 0 , .., (cid:22) n − (cid:23) = 2 m − − δ (2 m ) = ϕ (2 m )2 = 2 m − = n l + 1 = 2 2 m − + 1 , ..., n −
1, hencethere are n of them. With the cos formula given in the proof of proposition 4 ii) with p → n , they canbe rewritten in the notation, also used in the above context, when read backwards, as 1 , , ..., m − − n positive zeros, only the overall sign isdifferent. ii) For n ≥
3, not a power of 2, one uses PIE to count the positive zeros. For this consider the oddmultiples of some odd number a , say (2 k + 1) a , (only such multiples come up as 2 l + 1 values inthe product (20)) up to k max such that all positive zeros are reached. If the largest 2 l + 1 value in theproduct which leads to a positive zero is ¯ n , then k max = (cid:22) (cid:16) ¯ na − (cid:17)(cid:23) , and there are k max + 1 of theseodd multiples of a . In the following application of PIE a will be taken as any odd prime or product ofodd primes from the set sqf kset ( n ). α ) Case even n , not a power of 2: The counting of the positive zeros starts at level r = 0 with allpositive ones in the unrestricted product eq. (20). There are (cid:22) n − (cid:23) + 1 of them (the l values are0 , , ..., (cid:22) n − (cid:23) ). In the next step, r = 1, all odd multiples of every odd prime p i in sqf kset ( n ) notexceeding 2 (cid:22) n − (cid:23) + 1 are discarded (there are M ( n ) such primes, where M ( n ) := | sqf kset ( n ) | − − M ( n ) X i =1 (cid:22) (cid:18) L ( n ) p i − (cid:19)(cid:23) + 1 with L ( n ) given in the proposition ii) α ). Now double subtractions may have appeared and one adds, at step r = 2, all odd multiples ofthe product of two odd primes from sqf kset ( n ), call them p i p i with i < i (because no problem ofover-subtraction appeared for i = i ). This leads to the term + X ≤ i
1) The counting of the positive zeros in the odd n case distinguishes the two cases 1 ( mod
4) and3 ( mod α ).One has just to determine the boundary values for the l max value leading still to a positive zero. This is l max = K −
1, if n = 4 K + 1, i.e. , n −
54 . Hence there are l max + 1 such l values to start with at level r = 0. Again the extra +1 can later be taken as r = 0 term (cid:0) M ( n ) (cid:1) for the vanishing alternating sumover row No. M ( n ) in Pascal ’s triangle. Here M ( n ) := | sqf kset ( n ) | . Therefore the PIE formula startswith the K ( n ) given in β
1) of the proposition . The L ( n ) of the PIE sum is now 2 n −
54 + 1 = n −
32 ,the largest 2 l + 1 value leading to a positive zero. β
2) In the case M ( n ) := | sqf kset ( n ) | , n = 4 K + 3 the maximal 2 l + 1 value leading to a positivezero, the L ( n ) in the formula, is 2 K + 1 = n −
12 , coming from the largest l value which is K . This K is the K ( n ) in the formula claimed for this case β ii) For the number of negative zeros the counting is done by finding all the odd multiples of odd primes,or products of them, which cover the 2 l + 1 range for the negative zeros. This is the difference of thenumber of such multiples for the whole range and the range for the positive zeros. α ) If n is even, not a power of 2, the whole range is determined by l max + 1 = n −
1. From the above ii) α ) case one knows the corresponding upper bound for the positive zeros, therefore the PIE formulahas N ( n ) = ( n −
22 + 1) − ( (cid:22) n − (cid:23) + 1), the number of factors in the unrestricted product whichlead to negative zeros, which is the value claimed in the proposition iii) α ). Accordingly, P ( n ) = n − L ( n ) = 2 (cid:4) n − (cid:5) + 1 as given in eq. (26). β
1) In this n ≡ mod
4) case the maximal number 2 l + 1 in the product is n −
2, determining P ( n ). The corresponding number L ( n ) for the positive zeros is taken from above as n −
32 . There are N ( n ) = (cid:18) n −
32 + 1 (cid:19) − (cid:18) n −
54 + 1 (cid:19) = n −
14 unrestricted factors in the product with negativezeros. β
2) In this n ≡ mod
4) case one has also P ( n ) = n − n . L ( n ) is taken from thecorresponding ii) case as n −
12 , and N ( n ) is obtained from (cid:18) n −
32 + 1 (cid:19) − (cid:18) n −
34 + 1 (cid:19) = n −
34 . (cid:3)
We give the first entries of the δ + ( n ) and δ − ( n ) sequences. Remember that the vanishing zero for n = 2is here counted as positive. The A-numbers given for δ − (2 k ) and δ − (2 k ) are conjectured. n even: δ + (2 k ) , k = 1 , , ... ; A055034 : [1 , , , , , , , , , , , , , , , , , , , , , , ... ], δ − (2 k ) , k = 1 , , ... ; A055034 with first entry 0 : [0 , , , , , , , , , , , , , , , , , , , , , , ... ], n odd, 1( mod δ + (4 k + 1) , k = 0 , , ... ; [1 , , , , , , , , , , , , , , , , , , , ... ], δ − (4 k + 1) , k = 0 , , ... ; [1 , , , , , , , , , , , , , , , , , , , , , ... ], n odd, 3( mod δ + (4 k + 3) , k = 0 , , ... ; [1 , , , , , , , , , , , , , , , , , , , , ... ], δ − (4 k + 3) , k = 0 , , ... ; [0 , , , , , , , , , , , , , , , , , , , , , ... ].The start of the sequences for odd n is therefore δ + (2 k + 1) , k = 0 , , ... ; [0 , , , , , , , , , , , , , , , , , , , , , , ... ], δ − (2 k + 1) , k = 0 , , ... ; [1 , , , , , , , , , , , , , , , , , , , , , , , , ... ].Finally, combining for all n : δ + ( n ) , n = 1 , , ... ; A193376 : [0 , , , , , , , , , , , , , , , , , , , , , , , , , ... ], δ − ( n ) , n = 1 , , ... ; A193377 : [1 , , , , , , , , , , , , , , , , , , , , , , , , , , ... ].On can check that δ ( n ) − ( δ + ( n ) + δ − ( n )) vanishes in each case.Observe that from these examples it seems that δ + (2 k ) = δ − (2 k ) = A055034( k ) = δ ( k ) for k ≥ δ (2 k ) = 2 δ ( k ), i.e. , ϕ (4 k )2 = 2 ϕ (2 k ) for these k values. The latter eq., whichholds also for k = 1, can be checked by considering the two cases k = 2 e (1) · (odd number) and k odd.Recall that δ (2) = δ (1) = 1, because δ (1) was special (it is not ϕ (2)2 ). Zeros of C ( n , x ) in the power basis < ρ ( n ) = , ρ ( n ) , ..., ρ ( n ) δ ( n ) − > As will be explained in the next section, the minimal polynomial C ( n, x ) with degree δ ( n ) leads to thesplitting field Q ( ρ ( n )) which is a simple field extension of the rational field Q , and the degree of Q ( ρ ( n ))12ver Q , the dimension of Q ( ρ ( n )), considered as a vector space over Q , is just the degree of C . In standardnotation [ Q ( ρ ( n )) : Q ] = δ ( n ) . Therefore it is interesting to write the zeros of C in the power basisfor this Q -vector space. This task is accomplished by using for the zeros ˜ ξ ( n ) l = cos (cid:16) πn (2 l + 1) (cid:17) , l ∈{ , , ..., (cid:22) n − (cid:23) } with gcd (2 l + 1 , n ) = 1 (see eq. (20)),the formula ˜ ξ ( n ) l = ˆ t (2 l + 1 , ρ ( n )) = 2 T (2 l +1 , ρ ( n )2 ) (compare with eq. (2)). Remember that the integer coefficient array for these ˆ t -polynomials isshown in A127672. Of course, one has to reduce this integer polynomials using C ( n, ρ ( n )) = 0, i.e. , onehas to replace ρ ( n ) δ ( n ) (and higher powers) by an integer polynomial of lower degree. This can be done ona computer and Table 4 has been found with the help of Maple13 [18]. For example, the zeros of C (15 , x )are ˜ ξ (15)1 = ρ (15) , ˜ ξ (15)2 = − ρ (15) + ρ (15) − ρ (15) , ˜ ξ (15)3 = − − ρ (15) + ρ (15) , ˜ ξ (15)4 =2 − ρ (15) . Here the reduction has been performed with ρ (15) → − ρ (15) + 4 ρ (15) + 4 ρ (15) − . . , − . − . ξ , ξ , ..., ξ δ ( n ) .Our next objective is to compute C ( n,
0) (the sign and the number multiplying x , also called the absoluteterm) and relate it first to the cyclotomic polynomial cy ( n, x ) (the minimal polynomial of the complexalgebraic number e π in , see e.g. , [30] or [9], p. 149, Exercise 50 a and b), evaluated at x = −
1, which is cy ( n, −
1) = ( − ϕ ( n ) n − Y k =1 gcd ( k,n )=1 (1 + e π i kn ) , n = 2 , , ... (29)We used the fact that the product in the definition of cy ( n, x ) has ϕ ( n ) factors (the degree as minimalpolynomial). For n = 1 one has cy (1 , −
1) = −
2, which fits this formula if one defines ϕ (1) := 1 andtakes the undefined product as 1. One may rewrite this for n ≥ e π in , using for the sumof the ϕ ( n ) terms the formula 2 n n − X k =1 gcd ( k,n )=1 k = ϕ ( n ) n , n = 2 , , ... (30)This is a formula listed by R. Zumkeller under A023022 which is found in[2], p. 48, exercises 15 and 16,written such that both sides depend only on the distinct primes in the prime number factorization of n .For the r.h.s. there is the well known formula [2], p. 27, ϕ ( n ) n = M ( n ) Y j =1 (cid:18) − p j (cid:19) = X d | n µ ( d ) nd , if n has the distinct prime factors p j , j = 1 , , ..., M ( n ) = A n ) = | sqf kset ( n ) | (see above for thenotion ‘squarefree kernel’). The M¨obius function µ entered here (see A008683). The PIE proof of eq.(30), given in the appendix A , uses this form of ϕ ( n ). If one uses also e i π = − cy ( n, −
1) = ( − ϕ ( n )2 n − Y k =1 gcd ( k,n )=1 (cid:18) π kn (cid:19) , n = 2 , , .... (31)This can also be related to a special Sylvester sequence. In general the complex cyclotomic
Sylvester -numbers Sy ( a, b ; n ) are related to the sequence with three term recurrence f n = a f n − + b f n − ,which has characteristic polynomial x − a x − b with zeros α ≡ α ( a, b ) = ( a + √ a + 4 b ) / β ≡ β ( a, b ) = ( a − √ a + 4 b ) /
2. The definition is (see e.g. , [19]) Sy ( a, b ; n ) := n − Y k =1 gcd ( k,n )=1 ( α − β e π i kn ) , n = 2 , , ... (32)13or n = 1 one takes Sy ( a, b ; 1) = α − β . For our purpose the values are ( a, b ) = (0 , − i.e. ,( α, β ) = ( i, − i ), and Sy (0 , − n ) := ( − ϕ ( n )2 n − Y k =1 gcd ( k,n )=1 (1 + e π i kn ) , n = 2 , , ..., (33)with Sy (0 , −
1; 1) = 2 i . This can be rewritten, using again eq. (30) and e i π = −
1, as Sy (0 , − n ) = ( − ϕ ( n ) n − Y k =1 gcd ( k,n )=1 (cid:18) π kn (cid:19) , n = 2 , , .... (34)Therefore Sy (0 , − n ) := ( − ϕ ( n )2 cy ( n, − , n = 2 , , .... (35)One could also include the n = 1 case, with ϕ (1) := 1 if one takes 1 i ϕ ( n ) instead of the prefactor.After these preliminaries back to the number C ( n, Proposition 6: C ( , ) , m ∈ N C (2 m,
0) = ( − ϕ (4 m )2 + ϕ (2 m ) Sy (0 , −
1; 2 m ) = ( − ϕ (4 m )2 + ϕ (2 m )2 cy (2 m, − , m ∈ N . (36) Proof:
Use eqs. (19) and (35). The number of factors in eq. (19) with n = 2 m is ϕ (4 m )2 , the degree δ (2 m ). Note that the restriction in the Sy (0 , −
1; 2 m ) product is gcd ( k, m ) = 1, while in the C (2 m, gcd ( k, m ) = 1, but this just says that only odd k s can contribute in both cases, and thoseodd numbers ≤ m − m or 2 m have to be omitted. Both restrictions exclude the same oddnumbers. (cid:3) Corollary 1: C ( , ), p an odd a prime For odd prime p : C (2 p,
0) = ( − p − p . (37) Proof:
This follows immediately from eqs. (19) and (31) if one uses the known formulae cy (2 p, x ) = cy ( p, − x ), cy ( p,
1) = p , and evaluates the ϕ functions to obtain the sign. This formula follows from cy ( n, x ) = Y d | n ( x d − µ ( nd ) where the multiplicative M¨obius function (with µ ( p ) = −
1) enters (see, e.g. , [9], exercise50 b., solution p.506).
Proposition 7: C ( m , x ) , m ∈ N C (2 m , x ) = ˆ t (2 m − , x ) , m ∈ N , (38)with the integer polynomials ˆ t defined in eqs. (2) and (3) in terms of Chebyshev T - or S − polynomials. Proof:
Use eq. (16) and the
Note added to the link [13], where it was shown in eq. (8) that2 m − Ψ (cid:16) m +1 , x (cid:17) = 2 T (cid:16) m − , x (cid:17) =: ˆ t (2 m − , x ), for m = 1 , , ... . It is clear that C (1 , x ) = x + 2.This proposition will later be generalized in theorem 1A . (cid:3) Corollary 2: C ( m , ) , m ∈ N From the proposition 7 and the known fact that T (2 n,
0) = ( − n follows that C (1 ,
0) = +2 , C (2 ,
0) = 0 , C (2 ,
0) = − , C (2 m ,
0) = +2 , m ≥ . (39)Note that a standard formula for the cyclotomic polynomials cy (2 m , x ) which involves divisors (see e.g. ,[9], p. 149, Exercise 50 b, with Ψ m ( x ) ≡ cy ( m, x )) leads to undetermined expressions for cy (2 m , −
1) if m ≥
1. 14or the following theorem 1A on a formula for C ( n, x ) for even n , based on the divisor product represen-tation ( dpr ) of numbers [14], we need two lemmata . Lemma 3: Pairing in dpr ( )In the divisor product representation of an even number 2 m , m ∈ N , for each a -factor in the numerator(resp. denominator) one finds exactly one a -factor in the denominator (resp. numerator) with argumentratio either 1 : 2 or 2 : 1. Proof:
Due to the ∗ -multiplication property of dpr s (see the theorem in [14]) it is sufficient to prove this lemma for dpr (2 p · · · p N ) with odd primes p , ..., p N . Indeed, if the prime factorization of 2 m is2 k +1 p e · · · p e N N , with k ∈ N and odd primes p j and positive exponents e j , for j = 1 , ..., N , then dpr (2 m ) = (2 k p e − · · · p e N − N ) ∗ dpr (2 p · · · p N ) and the pairing property of the latter dpr will be pre-served after multiplication of each argument. Now the proof of the pairing property for dpr (2 p · · · p N )becomes elementary, once one splits the products in eq. (7) of [14] (note that N in this reference is now N + 1) into those a -factors which are even ( i.e. , contain the even prime 2) and those which are odd ( i.e. ,those composed of only odd primes). Indicate a -factors with odd arguments, being the product of k oddprimes, by a o ( . ( k ) . ). Then any a -factor in the numerator (resp. denominator) with some even argument2 n is found in the denominator (resp. numerator) exactly once with the argument n . This is because foreach factor in the numerator product Π a (2 . ( N − j ) . ), with j ∈ { , ..., (cid:22) N (cid:23) } , say, a (2 p i · · · p i N − j ),there is exactly one factor in the denominator product Π a o ( . ( N − j ) . ), namely the one which usesthe same N − j odd primes a ( p i · · · p i N − j ). Similarly, for each a -factor in the denominator productΠ a (2 . ( N − (2 j + 1)) . ), with j ∈ { , ..., (cid:22) N − (cid:23) } , there is exactly one factor in the numerator productΠ a o ( . ( N − (2 j + 1)) . ). The first factor in the numerator, a (2 p · · · p N ) ( j = 0) is paired with the singleodd argument a -factor of the first product in the denominator Π a o ( . ( N ) . ) = a ( p · · · p N ) .Note, that the example for the primorial A002110(5) = 2310, given in [14], table O (falling arguments) in eq. (7) in [14]. See the fifth position in the N = 5 case there.Recall also, for later purposes, that because dpr (2 m ) has the same number of factors in the numeratorand in the denominator, viz N ( this balance holds true for all dpr ( n ), n ≥
2, due to proposition 5 in[14]), after the split into even and odd arguments the number of pairs with even argument in the numera-tor matches the one with even argument in the denominator. This will later lead to a balanced formula for C (2 m, x ) in terms of ˆ t -polynomials (the same number of ˆ t s in the numerator and in the denominator, viz N − ). (cid:3) Lemma 4: ˆ t ( n + 1 , x ) − ˆ t ( n − , x )ˆ t ( n + 1 , x ) − ˆ t ( n − , x ) = S ( n + 1 , x ) S ( n − , x ) = 2 T (cid:16) n , x (cid:17) = ˆ t (cid:16) n , x (cid:17) , for n ≡ mod . (40)This identity involves the integer polynomials ˆ t introduced earlier in eqs. (2) and (3). Proof:
Use the known identity for
Chebyshev polynomials, [20], p. 261, first line, specialized, T ( m + 1 , x ) − T ( m − , x ) = 2 ( x − U ( m − , x ), written for x replaced by x S ( m − , x ) = U ( m − , x t -polynomials from eq. (2). For n ≡ mod
4) this identity can be usedin the numerator as well as in the denominator. The n − independent factor ( x − x = ±
2. However, the lemma holds also for these x − values as can be seen after applying l’Hˆopital’srule, using the well known identity T ′ ( x, n ) = n U ( n − , x ). In the second to last step the well known15dentity, [20] p. 260, last line, written as 2 T (cid:16) m, x (cid:17) S ( m − , x ) = S (2 m − , x ) has been employed. (cid:3) Theorem 1A: C ( , x ) , m ≥
1, in terms of ˆt-polynomials
With the prime number factorization of 2 m = 2 k p e · · · p e N N with k ∈ N , odd primes p j and positiveexponents e j , j = 1 , ..., N , one has C (2 m, x ) = ˆ t (2 k − p e · · · p e N N , x ) Π ˆ t (2 k − ∗ ( N − ., x ) Π ˆ t (2 k − ∗ ( N − ., x ) · · · Π ˆ t (2 k − ∗ ( N − ., x ) Π ˆ t (2 k − ∗ ( N − ., x ) · · · . (41)with ˆ t (2 k − ∗ (0) ., x ) = ˆ t (2 k − p e − p e − · · · p e N − N , x ), and the products Π ˆ t (2 k − ∗ ( K ) ., x ) are over the (cid:0) N K (cid:1) factors with K primes from the set { p , p , ..., p N } multiplied by p e − p e − · · · p e N − N , i.e. , K -products of the form ˆ t (2 k − p e − p e − · · · p e N − N p i p i · · · p i K , x ). We used ∗ ( k ) . instead of . ( k ) . for theindices of the ˆ t polynomials to remind one of the extra factor (without the powers of 2 because the evenprime has been extracted everywhere) due to the ∗ -multiplication.Before we give the proof an example will illustrate this theorem . Example 3: k = , N = C (1350 , x ) = C (2 · , x ) = ˆ t (3 , x ) ˆ t (3 , x )ˆ t (3 , x ) ˆ t (3 , x ) = ˆ t (675 , x ) ˆ t (45 , x )ˆ t (135 , x ) ˆ t (225 , x ) . (42) Proof:
This is based on [14] applied for the minimal polynomials Ψ (cid:0) m, x (cid:1) (see the definition for C (2 m, x )given in eq. (16)). Let 2 k +1 p e · · · p e N N be the prime number factorization for 4 m . Only the squarefreekernel 2 p · · · p N is important due to the theorem in [14] for the divisor product representations ( dpr s)and the ∗ -multiplication. This means that one has to multiply after the computation of Ψ(2 p · · · p N ),using proposition 1 of [14], the first argument, the index in conventional notation, of each factor t ( n i , x )in the the numerator, and of each t ( m i , x ) in the denominator (see [14], eq. (1)) with the number2 k p e − p e − · · · p e N − N . Here the pairing lemma 3 is crucial which carries over to the indices of the t -factors in the numerator and denominator of [14], eq. (1). Because there is always at least one factor of2 in the number which multiplies every t index after the ∗ -multiplication, the pairing will occur alwaysbetween indices which are 0 ( mod
4) and 0 ( mod n even alternative in the definitionof t (cid:16) n, x (cid:17) from eq. (2) of [14], rewritten here as eq. (18), is relevant. Observe that the prefactorsin this definition of t (cid:16) n, x (cid:17) are not of interest, provided we take 2 (cid:16) T (cid:16) n , x (cid:17) − T (cid:16) n − , x (cid:17)(cid:17) which is the monic integer polynomial ˆ t (cid:16) n , x (cid:17) − ˆ t (cid:16) n − , x (cid:17) . See eq. (18), and the discussion inconnection with lemma 2 . C (2 m, x ) is written here as a rational function of monic integer polynomials.For each replaced t (cid:16) n, x (cid:17) . t (cid:16) n , x (cid:17) (either in the numerator or denominator, depending on where thelarger index appears), one can apply lemma 4 . This is how for each such t -quotient one obtains ˆ t (cid:16) n , x (cid:17) .Here the fact that the larger index of a pair is always equivalent to 0 ( mod
4) is important. From thestructure of the numerator and denominator of the original dpr (2 m ) with the separation of the evenand odd a -arguments (which carries over to the t -indices), discussed in the proof of lemma 3 , one nowfinds the numerators and denominators of the theorem. Just search the products for the even t -indices. E.g. , Π ˆ t (2 k − ∗ ( N − ., x ) in the numerator of the theorem originates from the quotient of productsΠ t (2 . ( N − ., x )Π t o ( . ( N − ., x ) before the ∗ -multiplication has to be applied. This leads, with eq. (18) and lemma4 , to Π ˆ t (cid:18) k +1 ∗ ( N − ., x (cid:19) after multiplying each ˆ t index in the product with 2 k p e − p e − · · · p e N − N ,where the ∗ reminds one to multiply each index with this number divided by 2 k .16s announced at the end of the proof of lemma 3 the number of ˆ t polynomials in the numerator isthe same as the one for the denominator, viz N − , which is due to the sum in the Pascal -triangle A007318row No. N over even, resp. odd numbered positions. (cid:3) Note that this theorem 1A , evaluated for x = 0 leads in general to undetermined expressions, remember-ing that ˆ t ( n,
0) = 0 if n is odd, and 2 ( − n if n is even. However, a correct evaluation (using l’Hˆopital ’srule) has to reproduce the result known from corollary 2 .The following factorization of the monic integer ˆ t -polynomials is related to theorem 1A . Theorem 1B: Factorization of ˆt-polynomials in terms of the minimal C-polynomials ˆ t ( n, x ) = Y d | op ( n ) C (2 n/d, x ) = Y d | op ( n ) C (2 k +1 d, x ) , (43)with op ( n ) = A000265( n ), the odd part of n , and 2 k is the largest power of 2 dividing n . The exponentsare k = k ( n ) = A007814( n ), k ∈ N .Before the proof we give an example. Example 4: n = − x − x + 35 x − x + x = ˆ t (10 , x ) = C (20 , x ) C (4 , x ) =( x − x + 19 x − x + 1) ( x − . (44) Proof:
This is modeled after a similar proof in [30].It is clear that both sides are monic integer polynomials. See the definition of ˆ t in eqs. (2), (3) and proposition 2 .In order to check the degree we consider the even and odd n cases separately. If n = 2 k ( n ) op ( n )with k ( n ) = A007814( n ) ≥ X d | op ( n ) δ (2 k +1 d ) = X d | op ( n ) ϕ (2 k +2 d )2 from the known degree δ of the C -polynomials, here for index >
1. Due to well known properties of the
Euler totient function (see e.g. ,[2],
Theorem
Theorem k ( n )+2 − k ( n )+1 op ( n ) = n ,the degree of ˆ t ( n, x ). In the odd case, if n = op ( n ) then X d | n ϕ (4 d )2 = X d | n d = n , again the degree ofˆ t ( n, x ).For the proof one compares the zeros of both sides. The zeros of ˆ t ( n, x ) are ˆ x ( n ) l = 2 cos (cid:16) (2 l + 1) π n (cid:17) for l = 0 , , ..., n −
1. This is known from the zeros of the
Chebyshev T -polynomials. The zerosof the C (2 k +1 d, x )-polynomials are used in the form given by eq. (20): 2 cos (cid:16) (2 l ′ + 1) π k +1 d (cid:17) for l ′ = 0 , , ..., k d −
1, where gcd (2 l ′ + 1 , k +1 d ) = 1, i.e. , gcd (2 l ′ + 1 , d ) = 1. Because the degreesmatch, it is sufficient to show that each zero of ˆ t ( n, x ) occurs on the r.h.s. . For n = 2 k ( n ) op ( n ), with k ( n ) = A007814( n ), consider gcd (2 l + 1 , n ) = gcd (2 l + 1 , k +1 op ( n )) = gcd (2 l + 1 , op ( n )) = g (someodd number). Hence 2 l +1 = (2 l ′ +1) g , with some l ′ and op ( n ) = d g , i.e. , d | op ( n ). Therefore 2 l + 12 n =2 l ′ + 12 k +1 d . Thus the ˆ t ( n, x ) zero 2 cos (cid:16) (2 l + 1) π n (cid:17) appears on the r.h.s. as one of the C (2 k +1 d, x ) zeros. (cid:3) . Remark 3: Derivation of Theorem 1A from Theorem 1B
Theorem 1B can be used as recurrence for the C -polynomials in terms of the ˆ t -polynomials. This issimilar to the case treated in [30] for the minimal polynomials Ψ of 2 πn (see A181875/A181876 for their17oefficients). The solution of this recurrence has been given in [14]. Indeed, theorem 1A has been derivedabove from this solution. Originally we found theorem 1B starting from theorem 1A building up iterativelya formula for ˆ t (2 k − p e · · · p e N N , x ) in terms of the C -polynomials. We give this formula as a corollary. Corollary 3: ˆt-polynomials in terms of C-polynomials ˆ t (2 k − p e · · · p e N N , x ) = e Y q =0 e Y q =0 · · · e N Y q N =0 C (2 k p q p q · · · p q N n , x ) , k = N , N ∈ N . (45)In order to find a simplified expression for C (2 m + 1 , x ) , m ∈ N we need the following lemma in orderto rewrite the quotient t ( even, x ) t ( odd, x ) , given the definition eq. (18). Lemma 5: for n = 2 M + 1 , M ∈ N one has t (cid:0) n, x (cid:1) t (cid:0) n, x (cid:1) = 2 M − n ˆ t ( n + 1 , x ) − ˆ t ( n − , x )ˆ t ( M + 1 , x ) − ˆ t ( M, x ) = 12 M +1 ( x − S ( n − , x )( x − S (2 M, √ x ) (46)= x + 22 M +1 S ( n − , x ) S ( n − , √ x ) . (47)Here new important monic integer polynomials enter the stage: Definition 1: q-polynomials
With n ∈ N define q ( n, x ) := S (2 n, x ) S (2 n, √ x ) . (48)That this defines indeed monic integer polynomials of degree n is shown by the next lemma . Lemma 6: q ( n, x ) = ( − n S (2 n , √ − x ) = S ( n, x ) − S ( n − , x ) . (49) Proof:
It is known from the o.g.f. of the
Chebyshev S -polynomials that the bisection yields S (2 n, y ) = S ( n, y −
2) + S ( n − , y − S (2 n , √ x + 2 ) = S ( n, x ) + S ( n − , x ). With x replaced by − x one has S (2 n, √ − x ) = ( − n ( S ( n, x ) − S ( n − , x )) which explains the second equation of the lemma .Therefore one has to prove S (2 n, x ) = S ( n, x ) − S ( n − , x ) . This identity can, for example, beproved using the o.g.f. for the square of the S -polynomials (for their coefficient table see A181878,also for the paper [15] given there as a link). This computation was based on the Binet - de Moivre formula for the S − polynomials. This o.g.f. was found to be 1 + z − z
11 + (2 − x ) z + z . The o.g.f. for { S (2 n, x ) } ∞ n =0 is (1 + z ) / (1 + (2 − x ) z + z ) (from the bisection). Because the o.g.f. for S ( n, x ) − S ( n − , x ) is (1 + z ) times the one for S ( n, x ) the completion of the proof is then obvious. (cid:3) Remark 4 : O.g.f. and coefficient array for the q-polynomials
From lemma 6 the o.g.f. Q ( z, x ) := ∞ X n =0 q ( n, x ) z n = (1 − z ) / (1 − x z + z ) from the known o.g.f. of the S -polynomials. This shows that the q ( n, x ) coefficients constitute a Riordan -array (infinite lowertriangular ordinary convolution matrix), which is in standard notation (cid:18) − x x , x x (cid:19) , meaningthat the o.g.f. of the column No. m sequence is 1 − x x (cid:18) x x (cid:19) m . This is the triangle A13077718here more information can be found. For example, the explicit form for the coefficients is Q ( n, m ) =( − n − m +12 (cid:18) n + m m (cid:19) if n ≥ m ≥ lemma 3 will also be used in the proof of the following theorem . Theorem 2A: C ( n , x ) , n ≥
3, odd, in terms of q-polynomials
With the prime number factorization of n = p e · · · p e N N with odd primes p , ..., p N and positive exponents e j , j = 1 , ..., N , one has C ( n, x ) = q (cid:0) n − , x (cid:1) Q i
This is analogous to the proof of theorem 1A . Again the dpr (2 n ) representation, from which onederives Ψ(2 n, x ), with the prime number factorization for the odd n , is considered. The ∗ -multiplicationproperty allows to consider dpr (2 p · · · p N ) with a subsequent multiplication of all arguments in thenumerator and denominator by p e − · · · p e N − N . Because of the paired numerator/denominator structuredue to lemma 3 one finds, either in the numerator or in the denominator quotients of the type t (2 k, x ) t ( k, x )which are up to a factor 2 k +12 ( x + 2) equal to q (cid:18) n − , x (cid:19) . Factors of powers of 2 are irrelevant(they have to cancel) because on both sides of theorem monic polynomials appear. The factors of x − q -polynomials in the numerator and denominator have also to match(see the remark at the end of the proof of lemma 3 ). This number is 2 N − . The structure of thenumerator of the theorem originates from products with the even indexed t (cid:16) k, x (cid:17) polynomials in thenumerator after ∗ - multiplication. Before this multiplication on has in the numerator t (cid:16) p · · · p N , x (cid:17) and t − products over all possibilities to leave out 2 , , ... of the odd primes from the set { p , ..., p N } . The ∗ -multiplication then leads to t (cid:16) n, x (cid:17) and t − products over all possibilities to divide n by these 2 , , ... odd primes. Together with the pairing partners from the denominator this leads to the q -polynomialsgiven in the numerator of the theorem . A similar argument produces the denominator q -polynomials.The number of these q − polynomials in the numerator, viz N − , matches the one for the denominator. (cid:3) Theorem 2B: Factorization of q-polynomials in terms of C-polynomials q ( n, x ) = Y < d | (2 n +1) C ( d, x ) , n ∈ N . (52) Example 6: n = q (17 , x ) = C (5 , x ) C (7 , x ) C (35 , x ) . (53)This checks. Proof: theorem 1B .It is clear that both sides are monic integer polynomials, and the degree fits due to n = X d | (2 n +1) ϕ ( d )2 − ϕ (1). See the properties of the Euler totient function mentioned above in the proof of theorem 1B .The zeros of q ( n, x ) are x ( n ) l = 2 cos (cid:18) π l + 12 n + 1 (cid:19) , for l = 0 , , ..., n −
1. This follows from definition 1 and lemma 6 with the zeros of S (2 n, √ − x ) which are known from those of Chebyshev S -polynomials.To show that each of these zeros appears on the r.h.s. for C ( d, x ) with some d | (2 n + 1) , d = 1, i.e. , as2 cos (cid:18) π l ′ + 1 q (cid:19) for some l ′ ∈ { , , ..., d − } and gcd (2 l ′ + 1 , d ) = 1 (see eq. 20), let gcd (2 l + 1 , n +1) = g , with some odd g with 2 n + 1 = d g . Then 2 l + 1 = (2 l ′ + 1) g = (2 l ′ + 1) n +1 d , and for each l ∈ { , , ..., n − } there is one l ′ ∈ { , , ... , d − } . (cid:3) Remark 5: Derivation of Theorem 2A from Theorem 2B
Theorem 2B can be used as recurrence for the C -polynomials in terms of the q -polynomials. See the remark 3 . We give the solution in the following corollary . Corollary 4: q-polynomials in terms of C-polynomials
With the prime number factorization of 2 n + 1 = p e · · · p e N N , with odd primes, n ∈ N , one has q ( n, x ) = e Y q =0 · · · e N Y q N =0 C ( p q · · · p q N N , x ) /C (1 , x ) . (54)The division by C (1 , x ) = x + 2 was necessary because not all q j -indices were originally allowed tovanish.These rational representations of C ( n, x ) do not lend itself to extraction of the value C ( n,
0) becauseundetermined 00 quotients appear. In the following we give the absolute term of C for prime indices. Proposition 8: C ( p , )For n = p , a prime, one has C ( p,
0) = − p − if p ≡ − p +14 if p ≡ Proof:
In eq. (31) with n = p there is no gcd -restriction on the product. Therefore, one can use aknown formula (see appendix B for a proof) n − Y k =1 (cid:18) π kn (cid:19) = ( − n − if n is odd,0 if n is even, (56)to obtain cy (2 , −
1) = 0 (which is also clear from the definition: cy (2 , x ) = x + 1), and for odd primes cy ( p, −
1) = ( − p − = +1. Because all k contribute in the product, one can use for even k the formulacos (cid:18) π Kp (cid:19) = − cos (cid:18) π p − Kp (cid:19) which shows that one generates again all odd k contributions, how-ever each with a minus sign. This leads to cy ( p, −
1) = ( − p − p − Y l =0 (cid:18) (cid:18) π l + 1 p (cid:19)(cid:19) = C ( p, .20hus C ( p,
0) = ±
1. The sign of C ( p,
0) is ( − δ + ( p ) from eq. (19) for n = p . This number δ + ( p ) ofpositive zeros has been found in proposition 5 to be p − if p ≡ mod p +14 if p ≡ mod C (2 ,
0) = 0, the sign is, of course, irrelevant. (cid:3)
We close this section with a conjecture on the discriminant of the C − polynomials. The discriminantof a monic polynomial p of degree n can be written as the square of the determinant of an n × n Van-dermonde matrix V n ( x ( n )1 , ..., x ( n ) n ) with elements ( V n ) i,j := ( x ( n ) i ) j , = 1 , ..., n and j = 0 , ..., n − x ( n ) i of p . Here the δ ( n ) zeros of C n are given in eq. (20). Another formula for thediscriminant (see e.g. , [26], Theorem 5.1, p. 218) is in terms of the derivative C ′ ( n, x ) and the ze-ros of C : ( − δ ( n ) ( δ ( n ) − δ ( n ) Y i =1 C ′ ( n, x ( n ) i ). The result is the sequence Discr ( C ( n, x )) = A193681( n ) =[1 , , , , , , , , , , , , , ... ] for n ≥
1. The following conjecture is on the se-quence q ( n ) = n δ ( n ) Discr ( C ( n, x )) which is A215041, [1 , , , , , , , , , , , , , , , , , , , , ... ]. Conjecture: explicit form of the q-sequenceo) q (1) = 1 (clear). i) If n = 2 k for k ∈ N then q ( n ) ! = 2. ii) If n = p k for odd prime p and k ∈ N then q ( n ) ! = p ( p k − +1) / . iii) if n = 2 k p ( i ) k i · · · p ( i N ) k iN with k ∈ N , the i j − th odd primes p ( i j ), where 2 ≤ i < i <... < i N , with N ∈ N if k = 0 and N ≥ k = 0, then q ( n ) ! = N Y j =1 p ( i j ) k − p ( i ) ki − ··· p ( i N ) kiN − P ( N,j ) , with P ( N, j ) = N Y l =1 ,l = j ( p ( i l ) − , = Y odd p | n p δ ( n ) p − . (57)The last eq. follows from the degree δ ( n ) = ϕ (2 n )2 . This last formula does, however, not work in thecases i) and ii) . One can compare this formula with the proven one for the discriminant of the cyclotomicpolynomials (the minimal polynomials of exp (2 π n ) (or any of the primitive n -th root of 1), as given in[25], eq. (1) p. 297. For this (slightly rewritten) formula see also A004124 and A193679. Example 7:ii) p = p (4) = 7, k = 3: q (7 ) = q (343) = 7 (7 +1) / = 1341068619663964900807. iii) n = 2 · · q ( n ) = 3 · · · · = 4316018525852839090954658176626149564980915348463203041.These values have been checked with the help of Maple13 [18]. Note added:
The conversion of the C polynomials in terms of Chebyshev S polynomials (A049310) hasbeen initiated by Ahmet Zahid K¨u¸c¨uk and was finished together with the author. See A255237 for thedetails. Q ( ρ ( n )) for C ( n , x ) , field extension and Galois group The algebraic number ρ ( n ) = 2 cos (cid:16) πn (cid:17) , n ∈ N , with minimal polynomial C ( n, x ) over Q of degree δ ( n ), has been studied in sect. 3. Each of these polynomials, being minimal, is irreducible. All rootshave been given in eq. (19) (or eq. (20)). C ( n, x ) is also separable because all of its roots are distinct.21ecause 2 cos (cid:18) π kn (cid:19) = ˆ t ( k, ρ ( n )) (see eq. (2), and the coefficient array A127672), with the monicinteger ˆ t -polynomials, each zero can be written as integer linear combination in the vector space basis < , ρ ( n ) , ρ ( n ) , ..., ρ ( n ) δ ( n ) − > , called the power basis. One has to reduce in ˆ t ( k, ρ ( n )) all powers ρ ( n ) p , p ≥ δ ( n ) with the help of the equation C ( n, ρ ( n )) = 0. See Table 4 for the zeros of C ( n, x ), n = 1 , ..., Example 8: n = The δ (8) = 4 zeros of C (8 , x ) = x − x + 2 are, with ρ ≡ ρ (8) = p √ ± ρ and ± ( − ρ + ρ ) = ± p − √
2. In this case the degree 4 coincides with the number of DSRs in the upper half planeand the negative real axis.This shows that the extension of the rational field Q , called Q ( ρ ( n )), obtained by adjoining just onealgebraic element (called a simple field extension) is the splitting field (Zerf¨allungsk¨orper in German)for the polynomial C ( n, x ). Note that even though the polynomial C ( n, x ) is from the ring Z [ x ] oneneeds Q ( ρ ( n )) with rational coefficients r j for the general element α = P δ ( n ) − j =0 r j ρ ( n ) j . For example, ρ (8) − = 2 ρ (8) − ρ (8) . Some references for field extensions and Galois
Theory are [6], [12], chpts. Vand VI, [3], and the on-line lecture notes [16]. The dimension of of Q ( ρ ( n )) as a vector space over Q is δ ( n ). This is the degree of the extension, denoted usually by [ Q ( ρ ( n )) : Q ], and it coincides with thedegree of the minimal polynomial for ρ ( n ). Of course, it is a proper extension only if δ ( n ) ≥ i.e. , for n ≥
4. This extension of Q is separable, i.e. , the minimal polynomial for the general algebraic number α given above is separable (has only distinct zeros). It is a normal field extension, meaning that everyirreducible rational polynomial with one root in Q ( ρ ( n )) splits completely over Q ( ρ ( n )). See, e.g., [6],Theorem 5.24, p. 108.We now consider a subgroup of the group of automorphisms of Q ( ρ ( n )), called A ut ( Q ( ρ ( n )), whichconsists of those elements σ which leave the subfield Q pointwise invariant (fixed point field Q ): σ : Q ( ρ ( n )) → Q ( ρ ( n )), α σ ( α ), with σ ( β ) = β for all β ∈ Q . The subgroup of A ut ( Q ( ρ ( n )) of theseso-called Q -automorphisms is called the Galois group of Q ( ρ ( n )) over Q , denoted by G al ( Q ( ρ ( n )) / Q ).Occasionally we abbreviate this with G n . In order to find the elements σ (we omit the label n ) of thissubgroup it is sufficient to know σ ( ρ ( n )), because of the usual rules for automorphisms: (i) σ ( α + β ) = σ ( α ) + σ ( β ), (ii) σ ( α β ) = σ ( α ) σ ( β ), and (iii) α = 0 ⇒ σ ( α ) = 0. Indeed, because of σ (1) = σ (1 ) = σ (1) σ (1), one has σ (1) = 1, and the images of products of ρ ( n ) are obtained from products of σ ( ρ ( n )).Applying σ on the equation C ( n, ρ ( n )) = 0 (minimal polynomial), leads to C ( n, σ ( ρ ( n ))) = 0, becausethe integer (rational) coefficients and 0 are invariant under the Q -automorphism σ we are looking for.Therefore we have exactly δ ( n ) distinct Q -automorphisms σ j , j = 0 , ..., δ ( n ) −
1, determined from thedistinct roots of C ( n, σ ( ρ ( n ))), viz σ j ( ρ ( n )) = ˜ ξ ( n ) j +1 with the zeros of C ( n, x ) ordered like in eq. (19) withincreasing k values (see the Table 4 ). By the same token C ( n, σ ( ˜ ξ ( n ) j )) = 0, for j = 1 , ..., δ ( n ) for every σ . Because all roots are distinct (separable C ) this leads to an isomorphism between G al ( Q ( ρ ( n )) / Q )and a subgroup of the symmetric group S δ ( n ) . The Q -automorphisms σ can therefore be identified withpermutations of the roots of C (see e.g. , [6] ch. 6.3, pp. 132 ff. One identifies the roots ˜ ξ ( n ) j with j , and σ j with the permutation (cid:18) ... δ ( n ) σ j (1) σ j (2) ... σ j ( δ ( n )) (cid:19) ≡ [ σ j (1) σ j (2) ... σ j ( δ ( n ))]. This subgroup of S δ ( n ) isAbelian because only products of powers of ρ ( n ) appear and due to the automorphism property of σ thiscarries over to the Galois group.
Example 9: n = Q -automorphisms α = r r ρ (5) with ρ (5) = ϕ , the golden section. In this case Q ( ϕ ) is as quadratic numberfield usually called Q ( √
5) with the basis < , ϕ > for integers in Q ( √
5) (see e.g. , [10], ch. 14.3, p.2207, where τ = ( ϕ − δ (5) = 2) Q -automorphisms are obtained from the solutions σ ( ϕ ) = ϕ or − ϕ = 1 − ϕ of C (5 , n ) = x − x −
1. Hence σ = id : σ (1) = 1 , σ ( ϕ ) = ϕ and σ : σ (1) = 1 , σ ( ϕ ) = 1 − ϕ . Because σ = σ , G is generated by σ , hence the Galois group G al ( Q ( ϕ ) / Q ) is the cyclic group Z (also known as additive group Z / Z or C , but we reserve C for theminimal polynomials). The fixed field for G is Q ; for the trivial subgroup with element σ it is Q ( ϕ ). Example 10: n = Q -automorphisms The three zeros ( δ (7) = 3) of C (7 , n ) are ˜ ξ (7)1 = ρ (7), ˜ ξ (7)2 = − − ρ (7) + ρ (7) and ˜ ξ (7)3 = 2 − ρ (7) (see Table 4 ). In the sequel we omit the argument 7. Computing powers of ρ modulo C (7 , ρ ) = 0 onefinds the Q -automorphisms σ = id , σ ( ρ = ˜ ξ ) = ˜ ξ , σ ( ˜ ξ ) = ˜ ξ , σ ( ˜ ξ ) = ˜ ξ and σ ( ρ = ˜ ξ ) = ˜ ξ , σ ( ˜ ξ ) = ˜ ξ , σ ( ˜ ξ ) = ˜ ξ . Therefore, the identification with S permutations is σ . = [1 2 3] = e , σ . = [2 3 1] and σ . = [3 1 2]. Each σ j permutation can be depicted in a circle diagram with verticeslabeled 1 , Z subgroup of S .We can also characterize these Galois groups G n by giving their cycle structure and depict them as cyclegraphs. For cycle graphs see, e.g. , [31] “Cycle graph” and “List of small groups” with all cycle graphs forgroups (also non-Abelian ones) of order 1 , ... ,
16. In order to manage powers of
Galois group elementswe first need some fundamental identities of
Chebyshev T -polynomials, whence ˆ t -polynomials.We will need iterations of ˆ t polynomials governed by the following well known identity. Lemma 7: Iteration of ˆt-polynomials ˆ t ( n, ˆ t ( m, x )) = ˆ t ( n m, x ) , n, m ∈ N . (58) Proof:
See [26], Exercise 1.1.6, p. 5, first with the trigonometric definition of
Chebyshev
T-polynomialswhich then carries over to the general polynomials defined by their recurrence relation. This identity isthen rewritten for ˆ t -polynomials. (cid:3) Lemma 8: mod n reduction of ˆt-polynomials in the variable ρ ( n )ˆ t ( k, ρ ( n )) = ( − ⌊ kn ⌋ ˆ t ( k ( mod n ) , ρ ( n )) , n ∈ N , k ∈ Z . (59)( − ⌊ kn ⌋ =: p n ( k ), the parity of (cid:4) kn (cid:5) , will become important in the following. Sometimes (cid:4) kn (cid:5) is calledquotient and denoted also by k \ n . Proof:
This follows, with k = l n + r , trivially from the trigonometric identity cos (cid:16) πn ( l n + r ) (cid:17) =cos (cid:16) π l + π rn (cid:17) = ( − l cos (cid:16) πn r (cid:17) , with r ∈ { , , ..., n − } . (cid:3) To simplify notation we will use also ˆ t n ( x ) for ˆ t ( n, x ). From lemmata e.g. , ˆ t ( t ( ρ (7)) =ˆ t ( ρ (7)) = − ˆ t ( ρ (7)) = − ( ρ (7) − t ( ρ (7)) from the identity cos (cid:16) πn ( n ± l ) (cid:17) = − cos (cid:16) πn l (cid:17) . This proves the following lemma . Lemma 9: Symmetry relation of ˆt polynomials ˆ t ( n − l, ρ ( n )) = − ˆ t ( l, ρ ( n )) , n ∈ N , l ∈ { , , ..., n } . (60) Lemma 7 which used the trigonometric ρ ( n ) definition can be rewritten as a congruence for the ˆ t -polynomials with indeterminate x . This is because all what is needed is that ρ ( n ) is a zero of C ( n, x ). Corollary 5: Congruence for ˆt polynomials modulo C -polynomials ˆ t ( k, x ) ≡ ( − ⌊ kn ⌋ ˆ t ( k ( mod n ) , x ) ( mod C ( n, x )) , n ∈ N , k ∈ Z . (61)An example will illustrate this before we give an another proof of this corollary based on known T -polynomial identities. 23 xample 11: Congruence for n =
7, k = ˆ t (9 , x ) = x − x + 27 x − x + 9 x and C (7 , x ) = x − x − x + 1. Polynomialdivision shows that ˆ t (9 , x ) = ( x + x − x − x + 9 x + 5 x − C (7 , x ) + ( − x + 2), henceˆ t (9 , x ) ≡ − ˆ t (2 , x ) ( mod C (7 , x )).The alternative proof of corollary 5 is based on the following factorization of Chebyshev S -polynomialsin terms of the minimal polynomials Ψ( n, x ) of cos (cid:18) πn (cid:19) over Q (for these polynomials see A049310and A181875/A181876). Proposition 9: Factorization of S-polynomials in terms of Ψ-polynomials S ( n − , x ) = 2 n − Y Start with the eq. (3) of [30], p. 471, written for n → n and for the ˆ t -polynomials instead ofthe T -polynomials. This is ˆ t ( n + 1 , x ) − ˆ t ( n − , x ) = 2 n +1 Y d | n Ψ (cid:16) d, x (cid:17) . Then use the identity [20], p.261, first line, with m → n , n → m and U -polynomials replaced by S − polynomials. This leads to theidentity ˆ t ( n + m, x ) − ˆ t ( n − m, x ) = ( x − S ( n − , x ) S ( m − , x ) , m ≤ n ∈ N . (63)Here we only need the case m = 1. Then dividing out 2 Ψ(1 , x ) = x − , x ) = x + 2 leadsto the claimed identity. (cid:3) To end the preparation for an independent proof of corollary 5 we state: Corollary 6: C-polynomial divides some family of S-polynomials C ( n, x ) | S ( l n − , x ) , n ≥ , l ∈ N . (64)This is clear from the definition of C in eq. (16) and the fact that 2 < n | l n , for l ≥ E.g. , C (10 , x ) = x − x + 5 divides the family { S (9 , x ) , S (19 , x ) , S (29 , x ) , ... } Proof of corollary 5: From the corollary 6 and the identity eq. (63) with n → l n and m → r one seesthat ˆ t ( k, x ) = ˆ t ( l n + r, x ) ≡ ˆ t ( l n − r, x ) ( mod C ( n, x )). Now ˆ t ( l n − r, x ) = ˆ t (( l − n + ( n − r ) , x ),and one can use l := l − r := n − r (remember that r ∈ { , , ..., n − } , hence r has identicalrange) as new l and r variables in this congruence, to get ˆ t ( l n + r , x ) ≡ ˆ t ( l n − r , x ) ( mod C ( n, x )) =ˆ t (( l − n + r, x ) ( mod C ( n, x )). This can be continued until one ends up withˆ t (0 n + k ( mod n ) , x ) ( mod C ( n, x )). (cid:3) Now we are in a position to compute powers of elements of the Galois group G n . The subset of oddnumbers 2 l + 1 < n entering the product in eq. (20) will be denoted by M ( n ). There are δ ( n ) (degreeof C ( n, x )) such odd numbers. Definition 2: The fundamental set M ( n ) M ( n ) := (cid:26) l + 1 (cid:12)(cid:12)(cid:12)(cid:12) l ∈ (cid:26) , ...., (cid:22) n − (cid:23)(cid:27) and gcd (2 l + 1 , n ) = 1 (cid:27) = { m ( n ) , ..., m δ ( n ) ( n ) } , (65)with m ( n ) = 1 and we use the order m i ( n ) < m j ( n ) if i < j . For n = 1 one takes M (1) = { } . Example 12: |M (2) | = δ (2) = 1 , M (2) = { } ; |M (14) | = δ (14) = 6 , M (14) = { , , , , , } .For M ( n ) see the row No. n of the array A216319.For later purpose we define here, for odd n , the extended fundamental set c M ( n ) and its first differenceset △ c M ( n ) Definition 3: The fundamental extended set c M ( n ) for odd n For n odd, ≥ c M ( n ) := { , m ( n ) = 1 , ..., m δ ( n ) ( n ) = n − , n + 2 } .24hus | c M ( n ) | = δ ( n ) + 2. Note that gcd ( n ± , n ) = 1 for odd n (proof by assuming the contrary: gcd ( n + 2 , n ) = d > n and n + 2 are odd. Then d | ( n + 2) and d | n , hence d | (( n + 2) − n ), d | 2, implying d = 2 or d = 1, but d > n thefirst difference set △ c M ( n ) = { , △ m ( n ) , ..., △ m δ ( n ) ( n ) , } with △ m j ( n ) := m j ( n ) − m j − ( n ), willbecome important later on. △ m ( n ) = 1 and △ m δ ( n )+1 ( n ) = 4 for each odd n .With definition 2 eq. (20) implies, σ j ( ρ ( n )) = σ j ( ˜ ξ ( n )1 ) = ˜ ξ ( n ) j +1 = 2 cos (cid:16) πn m j +1 ( n ) (cid:17) =ˆ t ( m j +1 ( n ) , ρ ( n )) for j ∈ { , , ..., δ ( n ) − } . (66) E.g. , n = 7: δ (7) = 3, M (7) = { , , } , σ ( ρ (7)) = ρ (7), σ ( ρ (7)) = ˆ t (3 , ρ (7)), and σ ( ρ (7)) =ˆ t (5 , ρ (7)).Because ˆ t is a rational integer polynomial σ j ( ρ ( n )) = σ j ( σ j ( ρ ( n ))) = ˆ t ( m j +1 ( n ) , σ j ( ρ ( n ))) = ˆ t ( m j +1 ( n ) , ˆ t ( m j +1 , ρ ( n ))) , (67)and with lemma 6 this becomes σ j ( ρ ( n )) = ˆ t (( m j +1 ( n )) , ρ ( n )). In general we have σ kj ( ρ ( n )) = ˆ t (( m j +1 ( n )) k , ρ ( n )) , for j ∈ { , , ..., δ ( n ) − } , n, k ∈ N . (68)Instead of powers of σ j ( ρ ( n )) we can therefore consider powers of m j +1 ( n ). Lemmata t polynomial with a product of elements from M ( n ) as its first argument (orindex) is again a ˆ t polynomial with first argument from M ( n ). In this way one can build sequences ofpowers, starting from any element of M ( n ). Trivially, 1 k = 1. Before proving this closure of M ( n )under powers, provided the rules for ˆ t polynomials are taken into account, we give two examples, andthen define a new equivalence relation on the integers, called ( M odd n), denoted by n ∼ . Example 13: Cycle structure for n = 12 (dodecagon) δ (12) = 4 , M (12) = { , , , } . 5 = 25 ≡ mod lemma 8 . The sign p n (25) in eq.(59) is here +. (If later the sign will be − , the mod n result will be underlined.) The first 2-cycle istherefore [5 , = 49 ≡ mod 12) (sign +), whence the second 2-cycle is [7 , = 121 ≡ mod 12) (sign +), producing the third 2-cycle [11 , n = 12entry in Table 6 . In this example the Galois group is not generated by one element. hence it is non-cyclic.In fact, G al ( Q ( ρ (12)) / Q ) = Z × Z = Z (see first entry in Table 8 ). The corresponding cycle graph isshown in Figure 4 as the first entry, where the shaded (colored) vertex stands for 1 and the open verticesshould here be labeled with 5 , (three 2-cycles). In this example it wasnot necessary to employ lemma 9 because the signs were always +. Example 14: Cycle structure for n = δ (7) = 3 , M (7) = { , , } . 3 ≡ mod − in eq. (59) is remembered bythe underlining. 2 / ∈ M (7), and now lemma 9 is used to rewrite this 2 as 7 − ∼ ∈ M (7) (or 3 ≡ M odd ∼ (or M odd 7) is used for the congruence in the newsense, due to lemmata · ≡ mod n ) (sign +). The first 3-cycle is therefore[3 , , M (7), and G al ( Q ( ρ (7)) / Q ) = Z , the cyclic group of order3 = δ (7). The corresponding cycle graph is a circle with three vertices, one of them, labeled 1, is shaded(colored) and the other two open ones are labeled by 3 and 5.This brings us to the definition of an equivalence relation n ∼ (or M odd n ) over the integers Z . Rememberthat the floor function for negative arguments is defined as ⌊− x ⌋ = − ⌊ x ⌋ if x ∈ N , and ⌊− x ⌋ = − ( ⌊ x ⌋ + 1) if 0 < x / ∈ N . Definition 4: Equivalence relation n ∼ on Z k, l ∈ Z , n ∈ N : k n ∼ l ⇔ a n ( k ) = a n ( l ), with the map a n : Z → I n := { , , ..., n − } , k a n ( k ),where a n ( k ) = r n ( k ) if p n ( k ) = +1 ,r n ( − k ) if p n ( k ) = − , (69)where we used the division algorithm to write k = q n ( k ) n + r n ( k ), with the quotient q n ( k ) ∈ Z andthe residue r n ( k ) ∈ I n . Note that q n ( k ) = (cid:22) kn (cid:23) . The sign p n ( k ) = ( − ⌊ kn ⌋ = ( − q n ( k ) correspondingto the parity of q n ( k ), appeared already in lemma 8 . a n (1) = 0 because − ≡ mod k n ∼ l we also write k ≡ l ( M odd n) (this should not to be confused with mod n ). Thereforethe sequence a n could also be called M odd n . The first of these 2 n -periodic sequences a n are found inA000007( n + 1) , n ≥ 0, (the 0-sequence), A000035, A193680, A193682, A203571, A203572 and A204453,for n = 1 , ..., 7, respectively.The smallest non-negative residue system mod n , viz , , ..., n − k modulo n Maple [18] uses r n ( k ) = modp ( k, n ). We also use k ( mod n ) for r n ( k ). The reader shouldverify that n ∼ is indeed an equivalence relation satisfying reflexivity, symmetry and transitivity. Thedisjoint and exhaustive equivalence classes are given by { [0] , [1] , .., [ n − } , called the smallest non-negative complete representative classes (or residue classes) M odd n (we omit the index n at the classes:[ m ] = n [ m ], written this way in order to distinguish this class from the ordinary one [ m ] n used in thearithmetic mod n ). These classes are defined by [ m ] = { l ∈ Z | l n ∼ m } = { l ∈ Z | a n ( l ) = m } . Because r n ( − k ) = 0 if k ≡ mod n ) and r n ( − k ) = n − r n ( k ) if k mod n ) (later listed as lemma 17 ) onecan characterize these residue classes also in the following way. Lemma 10: Complete residue classes Modd n l ∈ [0] ⇔ l ≡ mod n ) . Equivalently, [0] = [0] n ,l ∈ [1] ⇔ l ≡ mod n ) or ≡ − mod n ) . Equivalently, [1] = [1] n ∪ [2 n − n , ... l ∈ [ n − ⇔ l ≡ n − mod n ) or ≡ − ( n − 1) ( mod n ) . Equivalently, [ n − 1] = [ n − n ∪ [ n + 1] n . (70)For example, take n = 7, then [3] = [3] = [3] ∪ [14 − = { ..., − , − , , , ... } ∪{ ... − , − , , , ... } = { ..., − , − , − , − , , , , , ... } . If n = 2 one has [0] = [0] and [1] = [1] . The first differences in the class [0] are, of course, n , and in the class [ m ], for m = 1 , , ..., n − △ = 2 ( n − m ) and △ = 2 m , e.g. , n = 7 , m = 3 , △ = 8 , △ = 6 :3 , , , , , ... . This difference alternation invalidates certain theorems known for mod n . E.g. , Theorem 53 of reference [10], p. 50, is no longer true: take m = 5 , n = 7 , a = 3 , b = 17. a and b belong both to the class [3] as well as [3] but they do obviously not both belong to the class [3] . Thefollowing lemma 11 shows that is is sufficient to know the positive values, and append the negative ofthese values for each class [ m ], for m ∈ { , , ..., n − } . Lemma 11: Antisymmetry of the classes [ m ] , for m ∈ { , ..., n − } For n ∈ N and every m ∈ { , ..., n − } the elements of the equivalence class [m] are antisymmetricaround 0. Proof: This is obvious for the class [0] = { ... − n, − n, , n, n, ... } . For m > m ( mod n ) appears as a negative number of (2 n − m ) ( mod n ), and vice versa : the nega-tive of every positive numbers of [2 n − m ] n appears as a negative one of [ m ] n . This is true because − ( m + l n ) = (2 n − m ) − ( l + 1) 2 n , for every l ∈ N , and similarly, − ((2 n − m ) + l n ) = m − ( l + 1) 2 n , forevery l ∈ N . (cid:3) This leads immediately to the following corollary . 26 orollary 7: Non-negative elements of the classes [ m ]For n ∈ N and every m ∈ { , ..., n − } one has[ m ] ≥ = [ m ] n, ≥ ∪ [2 n − m ] n,> , and [ m ] = [ m ] ≥ ∪ − ([ m ] > ) . (71)Here we used the notations [ m ] ≥ and [ · ] n, ≥ or [ · ] n,> to denote the subset of non-negative numbersof [m] and the non-negative or positive numbers of the ordinary residue classes mod n , respectively. Ofcourse 0 appears only in the class [0]. − ([ m ] > ) is obtained from the set [ m ] > (excluding 0 in the case ofclass [0]) by taking all elements negative. Note that [ − m ] is not used here.With g ( n, m ) := gcd ( m, n − m ) one has, for m ∈ { , , ..., n − } , [ m ] n,> = g ( n, m ) [ m/g ( n, m )] n/g ( n,m ) ,> and [2 n − m ] n,> = g ( n, m ) [(2 n − m ) /g ( n, m )] n/g ( n,m ) ,> . This is. of course, only of interest if g ( n, m ) = 1. E.g. , n = 6 , m = 3 , g ( n, m ) = 3: [3] ,> = 3 ∗ [1] ,> and [9] ,> = 3 ∗ [3] ,> , whereagain k ∗ [ p ] q,> is the set with all members of the set [ p ] q,> multiplied with k . This is obvious.The trivial formula for the members of the residue classes [ m ] ≥ , considered as sequences of increasingnumbers called { c ( n, m ; k ) } ∞ k =1 , is c ( n, m ; k ) = ( k − n if m = 0 , (cid:4) k (cid:5) n + ( − k m if m ∈ { , , ..., n − } . (72)The nonnegative members of the complete residue classes ( M odd n) for n = 3 , , , , and 7 are found inA088520, A203575, A090298, A092260, and A113807. Sometimes 0 has to be added, in order to obtainthe class [0]. Of course, these complete residue classes can be recorded as a permutation sequence of thenon-negative integers.We now list several lemmata (the trivial lemma 17 has already been used) in order to prepare for theproof of the multiplicative structure of these equivalence classes. Lemma 12: Parity of Modd n residue classes For even n the parity of the members of the residue class n [ m ], m ∈ { , , ..., n − } coincides with theone of m . If n is odd this is also true for the classes with m ∈ { , , ... , n − } , and for m = 0 the parityof the elements alternates, starting with + (for even). Proof : The case of the residue class [0] is clear for even or odd n because of its members 0 mod n (see lemma 10 ). Similarly, for the other m values, because then mod n applies. Lemma 13: Periodicity of the parity sequence p n with period length 2 n p n ( k ) = p n ( k + 2 n l ) for l ∈ Z , i.e. , p n ( k ) = p n ( r n ( k )) = p n ( k ( mod n )) , for k ∈ Z , n ∈ N . (73) Proof: This is obvious from the definition of p n ( k ) given in lemma 8 , eq. (59), with definition 4 , eq. (69).For the second part use k = 2 n q n ( k ) + r n ( k ). (cid:3) Lemma 14: (A)symmetry of sequence p n around k = for n ∈ N , k ∈ N : p n ( − k ) = + p n ( k ) if k ≡ mod n ) − p n ( k ) , if k mod n ) . (74) Proof: This follows immediately from the property of the ⌊− x ⌋ function mentioned above before defini-tion 4 . (cid:3) Lemma 15: Product formula for the residue r n For n ∈ N and k, l ∈ Z one has: r n ( k l ) = r n ( k ) r n ( l ) ( mod n ). Proof: Just multiply k = q n ( k ) n + r n ( k ) with l = q n ( l ) n + r n ( l ). (cid:3) Lemma 16: Residue r from r n For n ∈ N and k ∈ Z one has: r n ( k ) = r n ( k ) iff r n ( k ) ∈ { , , ..., n − } , r n ( k ) + n iff r n ( k ) ∈ { n, n + 1 , ..., n − } . Proof: Obvious for r n ( k ) ∈ { , ..., n − } as well as ∈ { n, ..., n − } . (cid:3) emma 17: Residue for negative numbers For n ∈ N and k ∈ N one has: r n ( − k ) = k ≡ mod n ) , n − r n ( k ) if k mod n ) . Proof: Obvious. (cid:3) Lemma 18: Symmetry of the a n (or Modd n) sequence around k = For n ∈ N and k ∈ N one has: a n ( − k ) = a n ( k ). Proof: This is proved for the two cases k ≡ mod n ) and k mod n ) separately. In the first case r n ( k ) = 0, as well as r n ( − k ) = 0 , hence a n ( k ≡ mod n )) = 0, which is symmetric around k = 0. Inthe other case, we employ lemma 13 , noting that r n ( k ) = 0 , n (otherwise k ≡ mod n )). The two cases r n ( k ) ∈ { , , ..., n − } and r n ( k ) ∈ { n + 1 , n + 2 , ..., n − } have p n ( k ) = p n ( r n ( k )) equal +1 or − 1, respectively. Then with the second alternative of lemma 14 one finds a n ( − k ) = r n ( − k ) if p n ( − k ) = − p n ( k ) = +1 and a n ( − k ) = r n ( k ) if p n ( − k ) = − p n ( k ) = − a n (+ k ). (cid:3) Now we turn to the arithmetic structure of the M odd n residue classes. It is clear from the followingcounter-example that addition cannot be done class-wise. Consider n = 6, k = 2 and l = 7. Then a (2 + 7) = 3 but a ( a (2) + a (7)) = a (2 + 5) = a (7) = 5. In other words, 2 6 ∼ ∼ 5, but2 + 7 = 9 6 ∼ ∼ 5, and 3 is not equivalent to 5 ( M odd a − n ∼ a n ∼ 1, because for n > a can also be of the form − k n if it belongs to a class [ m ] with positive m , whereas in the first case it has to be of the form 1 + k ′ n ,which can only match for n = 1 and 2. However, it turns out that multiplication can be done class-wise.This is the content of the following proposition . Proposition 10: Modd n residue classes are multiplicative For n ∈ N and k, l ∈ Z one has: a n ( k l ) = a n ( a n ( k ) a n ( l )) i.e. , k l n ∼ a n ( k ) a n ( l ) , i.e., kl ≡ a n ( k ) a n ( l ) ( M odd n ) . (75)Before giving the proof, consider the example n = 6, k = 2 and l = 7. Now a (2 · 7) = 2 and a ( a (2) · a (7)) = a (2 · 5) = a (10) = 2. Or stated equivalently, 2 · ∼ ∼ · ∼ a (2) a (7). Proof:i) Due to the symmetry of a n (see lemma 18 ) it is clear that it is sufficient to consider only non-negative k and l . ii) Consider first the cases k ≡ mod n ) or l ≡ mod n ). If m ≡ mod n ) then a n ( m ) = 0. Thisfollows for both alternatives in eq. (69). Therefore, if k ≡ mod n ), a n ( k l ) = 0 because k l ≡ mod n )for every l , and a n (0 · a n ( l )) = a n (0) = 0, proving the assertion. In the other case, l ≡ mod n ), theproof is done analogously. iii) Now r n ( k ) and r n ( l ) are non-vanishing, and k and l are positive. Four cases are distinguishedaccording to the signs of ( p n ( k ) , p n ( l )), viz (+ , +), ( − , − ), (+ , − ) and ( − , +). (+,+) : In this case, due to the 2 n -periodicity of p n (see lemma 13 ), r n ( k ) and r n ( l ) are both from { , , ..., n − } , hence (+ , +) : r n ( k ) = r n ( k ) a nd r n ( l ) = r n ( l ) . (76)Also, from eq. (69), one obtains in this case a n ( k ) a n ( l ) = r n ( k ) r n ( l ). There is an alternative for a n ( r n ( k ) r n ( l )), depending on p n ( r n ( k ) r n ( l )) being +1 or − 1. Both cases are possible as the followingexamples for n = 6 show: p (3 · 4) = p (12 ( mod p (0) = +1 and p (3 · 7) = p (21 ( mod p (9) = − 1. First the argument of p n is rewritten with the help of eq. (76). p n ( r n ( k ) r n ( l )) = p n ( r n ( k ) r n ( l )). Due to the (2 n )-periodicity ( lemma 13 ) this is p n ( r n ( k ) r n ( l ) mod (2 n )) = p n ( r n ( k l )) =28 ( k l ), due to lemma 15 with n → n , and again the (2 n )-periodicity. In the first alternative, the +1 case, a n ( a n ( k ) a n ( l )) = r n ( k ) r n ( l ) ( mod n ) = r n ( k l ), again from lemma 15 . This is just a ( k l ) if p n ( k l ) =+1, proving the assertion for this alternative. In the other case, p n ( k l ) = − a n ( r n ( k ) r n ( l )) = − ( r n ( k ) r n ( l ))( mod n ) which is rewritten with mod n -arithmetic and lemma 15 as − r n ( k l )( mod n ) = − ( k l )( mod n ) = r n ( − k l ). This coincides with a ( k l ) for this alternative, proving the assertion. (-,-) : Now we have from lemma 16 ( − , − ) : r n ( k ) = n + r n ( k ) a nd r n ( l ) = n + r n ( l ) . (77)Here p n (( r n ( − k ) r n ( − l )) is rewritten with lemma 17 , eq. (77), the 2 n periodicity, and lemma 15 ,as follows. p n (( n − r n ( k )) ( n − r n ( l ))) = p n ((2 n − r n ( k )) (2 n − r n ( l ))) = p n ( r n ( k ) r n ( l )) = p n ( r n ( r n ( k ) r n ( l ))) = p n ( r n ( k ) r n ( l ) ( mod n )) . With lemma 15 (with n → n this becomes p n ( r n ( k l )) = p n ( k l ), again from the (2 n )-periodicity. In the first alternative p n ( k l ) = +1 and a n ( r n ( − k ) r n ( − l )) = r n ( r n ( − k ) r n ( − l )). With lemma 17 and mod n -arithmetic this is r n (( − r n ( k )) ( − r n ( l ))) = r n ( r n ( k ) r n ( l )), and with lemma 15 this becomes r n ( k l ), coinciding with a n ( k l ) for this alternative. Forthe other alternative, p n ( k l ) = − a n ( r n ( − k ) r n ( − l )) = − ( r n ( − k ) r n ( − l ))( mod n ). This becomes, with lemma 17 , mod n -arithmetic and lemma 15 − ( r n ( k ) r n ( l )) ( mod n ) = − r n ( k l )( mod n ). This vanishes if r n ( k l ) = 0 which means k l ≡ mod n )(which is possible, e.g. , n = 6 , k = 2 , l = 3), and then this coincides with the claim which is for thisalternative a ( k l ) = r n ( − ( k l )) = 0. If r n ( k l ) = 0 then − r n ( k l ) ( mod n ) = n − r n ( k l ) which alsocoincides with the claim a ( k l ) = n − r n ( k l ) if k l mod n ).We skip the proofs of the other two cases, (+ , − ) and ( − , +), which run along the same line. Here onearrives first at p n ( − ( k l )), and in order to compare it with p ( k l ) both alternatives in lemma 17 have to beconsidered like in the just considered second alternative. (cid:3) For the computation of the cycle structure of the Galois group G n = G al ( Q ( ρ ( n )) / Q ) we are onlyinterested, due to eq. (20), in odd numbers relatively prime to n . Contrary to ordinary mod n -arithmeticwhere the set of odd numbers O := { l + 1 | l ∈ Z } is in general not closed under multiplication( e.g. , 5 · ≡ mod O is closed under M odd n multiplication. Ofinterest are the units, the elements which have inverses, in order to see the expected group structure.First consider the reduced set O ∗ n , given by the odd numbers relatively prime to n . The negative oddnumbers in this set are just the negative of the positive odd numbers, therefore it will suffice to consider O ∗ n,> := { l + 1 | l ∈ N , gcd (2 l + 1 , n ) = 1 } . If n is a power of 2 the set O ∗ will be O , with the o.g.f. G ( x ) = x (1 − x ) (1 + x ) for the sequence of positive odd numbers { o ∗ ,> ( k ) := 2 k + 1 } ∞ k =0 . For theother even numbers n only the odd numbers relatively prime to the squarefree kernel of n , called sqf k ( n ),(see A007947, encountered already several times), will enter the discussion. Therefore, besides the justconsidered (trivial) case of the even prime 2, the set O ∗ n is only relevant for squarefree odd moduli, eitherprime or composite. We consider first the case of odd primes n = p , and give the o.g.f. of the sequencesof numbers from O ∗ p,> , called o ∗ p,> ( k ), as well as an explicit formula in terms of floor functions. Theseare the positive odd integers without odd multiples of the odd p . Proposition 11: Odd prime moduli, o.g.f. and explicit formula for O ∗ p ,> elements With G p ( x ) := ∞ X k =0 o ∗ p,> ( k ) x k , for odd primes p , one has G p ( x ) = x (1 − x p − ) (1 − x ) p − X k =1 x k (1 + x p − ) + 4 p − + x p − , (78)29nd o ∗ p,> ( k ) = k = 0 ,2 k − (cid:22) k + p − p − (cid:23) if k ∈ N . (79)We have used the even number 0 for k = 0 such that the G p ( x ) sum can start with k = 0. Proof : This proposition will become a corollary to the later treated general case of odd squarefree moduli n in proposition 13 . Example 15: O.g.f.s and formula for reduced odd numbers for modulus p = G ( x ) = x (1 − x ) (1 − x ) (cid:8) x + x ) + 4 x + 2 ( x + x ) + x (cid:9) , (80) o ∗ ,> ( n ) = 2 n − (cid:22) n + 26 (cid:23) , n ≥ . (81)The instances for p = 3 , , , , 13 and 17 are found under A007310, A045572, A162699, A204454,A204457, and A204458, respectively.In order to prepare for the general case of odd squarefree modulus n , we state a proposition on thestructure of the reduced odd numbers set O ∗ n,> . Proposition 12: Mirror symmetry and modular periodicity of O ∗ n ,> for odd ni) Mirror symmetry. For k ∈ { , , ..., δ ( n ) } one has: o ∗ n,> (2 δ ( n ) − ( k − n − o ∗ n,> ( k ) , (82)where o ∗ n,> ( k ) = m k ( n ) from M ( n ), given in definition 2 , eq. (65). ii) mod n periodicity: For k ∈ N one has: o ∗ n,> ( k ) = o ∗ n,> ( k + 2 δ ( n )) ( mod n ) . (83)Written as a relation between neighboring fundamental units, numbered by N ≥ 1, this becomes thefollowing statement. For k ∈ { N − δ ( n ) + 1 , ..., N δ ( n ) } , with N ∈ { , , , ... } , one has: o ∗ n,> ( k ) = ( N − 1) 2 n + o ∗ n,> ( k − ( N − 1) 2 δ ( n )) . (84) δ ( n ) is the degree of the minimal polynomial C ( n, x ) for the algebraic number ρ ( n ) introduced in section 2 .Note that if n = ω ( n ) Y j =1 p j with distinct odd primes p , and ω ( n ) = A001221( n ), then 2 δ ( n ) = ω ( n ) Y j =1 ( p j − L ( n ) := 2 δ ( n ) is the length of the fundamental N -units.Before we give the proof consider figure 3 for the case n = 3 · δ ( n ) = 4. The secondstatement ii) concerns the relation of the numbers of the second fundamental unit ( N = 2) to the onein the N = 1 unit. E.g. , the odd number 37 for k = 2 · o ∗ ,> (10 − · 4) =30 + o ∗ ,> (2) = 30 + 7, which checks. The statement i) shows the mirror symmetry within the first(and any other) unit of length L (15) = 2 · figure 3 this symmetry is indicated by the bracketsbelow the first unit, and it is a symmetry around the missing number n = 15. Missing numbers havebeen indicated by a dot. It is the pattern of missing odd numbers which is mirror symmetric, not theone of the actual values of the odd numbers. But the relation between the odd numbers in the secondhalf of a unit and the first one then follows, and is given by the statement of the proposition . In figure3 P ( n ) := 2 n = 30 is the shift for the o ∗ >,n values from the N = 1 to the N = 2 unit (or any of theneighboring units), and p ( n ) := n + 1 = 16 is the shift for these values from the first half of every unitto the second half. 30 * (k) >,n 2 d (n) d (n)4 +p(n)=n+1=16 + P(n) = 2n = 30 d (n) d (n)3 N=1 N=2 ...... k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 n−2 .. n+2 Figure 3: Structure of the sequence of mod n reduced odd numbers, n=15 Proof: i) It is clear from the degree of the minimal polynomial and eq. (20) that the number of mod n reduced positive odd numbers smaller than 2 n is δ (2 n ) = 2 δ ( n ) (from the definition of δ ( n ) in terms of Euler ’s ϕ function). We now determine δ ( n ) reduced odd numbers mod n which lie between n and 2 n ,by mirroring the δ ( n ) elements of M ( n ) around the position where the missing number n is situated,which is between the position k = δ ( n ) and the next one. The mirror symmetry refers to the gaps. Thenumber m k ( n ) from M ( n ) will be mapped to o ∗ n,> (2 δ ( n ) − ( k − N = 1). To find this odd number the first differenceset △ c M ( n ), introduced in connection with c M ( n ) from the definition 3 , becomes important in order tocount the gaps in the sequence of odd numbers when they are reduced mod n . o ∗ n,> (2 δ ( n ) − ( k − m k ( n ) twice the value of the sum of the gaps from the position k to the center, themirror axis. This is 2 ( △ m k +1 ( n ) + △ m k +2 ( n ) + ... + △ m δ ( n ) ( n ) + 2). The 2 in this sum is half thegap-length from the value n − δ ( n ) and n + 2 at the next position, the mirror-position δ ( n ) + 1 of δ ( n ). This is a telescopic sum which becomes 2 ( − m k ( n ) + m δ ( n ) ( n ) + 2). Therefore, thevalue of o ∗ n,> (2 δ ( n ) − ( k − m k + 2 ( − m k ( n ) + m δ ( n ) ( n ) + 2) = 4 + 2 m δ ( n ) ( n ) − m k ( n ) =2 n − m k ( n ), because always m δ ( n ) ( n ) = n − figure 3 , n = 15 with the values 30 − − − 11 = 19 and 30 − 13 = 17 for k = 1 , , N = 1 unit, because gcd ( m k ( n ) , n ) = 1, as member of M ( n ), implies for the mirror imagealso gcd ( o ∗ n,> (2 δ ( n ) − ( k − , n ) = 1. All positive odd numbers relatively prime to n and not exceeding2 n − m ( n ) = 2 n − ii) : From above we know that the number at position k = 2 δ ( n ) + 1 is 2 n + 1 because theone for k = 2 δ ( n ) has been shown to be 2 n − gcd (2 n + 1 , n ) = 1 (indirect proof by assuming thecontrary, using odd n ; this is similar to the proof given in connection with definition 3 of c M ( n )). Now itis clear that a shift in the o ∗ n,> numbers with P ( n ) := 2 n leads from the fundamental unit No. N = 1to the second one, N = 1, by putting o ∗ n,> (2 δ ( n ) + k ) = o ∗ n,> ( k ) + P ( n ), for k ∈ { , , ..., δ ( n ) } .This is obvious because the gcd value 1 is not changed by adding 2 n . This process can be iteratedto find the mod n periodicity structure stated in ii) . See figure 3 , n = 15 with k = 14, N = 2: o ∗ ,> (14) = 49 = 2 · 15 + o ∗ ,> (14 − · · 4) = 30 + o ∗ ,> (6) = 30 + 19. (cid:3) The sequences { o ∗ n,> } for n = 15 and n = 21 are found in A007775 and A206547, respectively.The sequences of the M odd n residues of the numbers o ∗ n,> ( k ), for k = 1 , , ... , for prime moduli n = p = 3 , , , , , 17, and for the first odd composed ones for n = 15 , 21 are found in A000012,A084101, A110551, A206543, A206544, A206545, and A206546, A206548, respectively.31or the following formula for the nonnegative odd numbers reduced mod n involving floor functions weneed the following list (increasingly ordered set) F ( n ) of length L ( n ) = 2 δ ( n ) derived from the list △ c M . Definition 5: List F ( n ) F ( n ) = { f ( n, , ..., f ( n, δ ( n ) } with f ( n, j ) = △ m j +1 ( n ) − , for j ∈ { , , ..., δ ( n ) − } f ( n, δ ( n )) = 1 , f ( n, δ ( n ) + j ) = f ( n, δ ( n ) − j ) , for j ∈ { , , ..., δ ( n ) − } , and f ( n, δ ( n )) = 0 . (85)Here △ m k ( n ) := m k ( n ) − m k − ( n ), (the k th element of △ c M ( n )). This list F ( n ) is obtained from firstenlarging △ c M by mirroring the first δ ( n ) entries at the last element 4, to obtain a list of order 2 δ ( n ) + 1.Then the first and last element of this new list is put to zero and all the other elements are diminishedby 2, thus obtaining a set of only even numbers. Then one divides by 2, omits the final 0, and reversesthe remaining list. Example 16: F ( ) n = 15 = 3 · 5, 2 δ (15) = 2 · M (15) = { , , , } , c M (15) = { , , , , , } , △ c M (15) = { , , , , } , the mirror extension is { , , , , , , , , } , the reduction step leads to { , , , , , , , , } ,and finally, dividing by 2, omitting the last 0 and reverting, leads to F (15) = { , , , , , , , } . Exceptfor the last 0 there is a mirror-symmetry around the fourth entry 1. See the first row of table 5 .For the odd squarefree composite numbers n = A024556( m ), m = 1 , , ... , 17, see table 5 for c M ( n ), △ c M ( n ) and F ( n ). Proposition 13: O.g.f. and formula for O ∗ n ,> elements. With G n ( x ) := ∞ X k =1 o ∗ n,> ( k ) x k , for odd n , one has G n ( x ) = x (1 − x δ ( n ) ) (1 − x ) δ ( n ) − X k =1 △ o ∗ n,> ( k + 1) x k + x δ ( n ) (86)with the first differences △ o ∗ n,> ( j ) := o ∗ n,> ( j ) − o ∗ n,> ( j − 1) .This generates o ∗ n,> ( k ) = 2 k − δ ( n ) X j =1 f ( n, j ) (cid:22) k + ( j − δ ( n ) (cid:23) . (87)Note that the numerator polynomial in G n ( x ) needs the first differences of the sequence members of thefundamental N = 1 unit, which due to the mirror symmetry of proposition 12 can be reduced to thefirst differences of { m ( n ) , m ( n ) , .., m δ ( n ) , n + 2 } with m δ ( n ) = n − proposition 11 , o n,> (0) = 0 and let the sum in G n ( x ) start with k = 0. Proof: We start with the periodicity of the sequence due to proposition 12 ii) . Taking the difference o ∗ n,> ( k ) − o ∗ n,> ( k − 1) = o ∗ n,> ( k − δ ( n )) − o ∗ n,> ( k − − δ ( n )) leads to the recurrence (we omitall unnecessary indices) o ( k ) = o ( k − 1) + o ( k − δ ( n )) − o ( k − − δ ( n )), for k ≥ δ ( n ) + 1.Here one needs also o ∗ n,> (0) := − 1. For some of these recurrences the o.g.f. s were determined by R. J. Mathar in e.g. , A045572 and A162699. I general G n ( x ) = δ ( n ) X k =1 o ( k ) x k + x ∞ X k =2 δ ( n )+1 o ( k − x k − + x δ ( n ) ∞ X k =2 δ ( n )+1 o ( n − δ ( n )) − x δ ( n )+1 ∞ X k =2 δ ( n )+1 o ( n − δ ( n ) − x n − δ ( n ) − , where the32ecurrence has been used (and the infinite sum has been reordered, not bothering about absolute con-vergence, in the sense of formal power series). Shifting the indices, one arrives at G n ( x ) = δ ( n ) X k =1 o ( k ) x k + x n G n ( x ) − ( o (1) x + .... + o (2 δ ( n ) − x δ ( n ) − o + x δ ( n ) G n ( x ) − x δ ( n )+1 ( G n ( x ) + ( − − o (0) = − − x − x δ ( n ) + x δ ( n )+1 ) G n ( x ) = x (cid:0) △ o (2) x + ... + △ o (2 δ ( n )) x δ ( n ) − + x δ ( n ) (cid:1) . Factorizing the bracket onthe l.h.s. and division leads to the claimed form for G n ( x ). Of course, the denominator factor (1 − x δ ( n ) )can be factorized into cyclotomic polynomials.For the proof of the second part we use L := 2 δ ( n ). Given G n ( x ) with the L + 1 input coefficients of thenumerator polynomial one derives the claimed explicit form for o ∗ n,> ( k ), by defining first the sequence withentries b L ( k ) := (cid:22) k + L − L (cid:23) , generated by x (1 − x L ) (1 − x ) (partial sums of the characteristic sequencefor multiples of L , then shifted). The numerator polynomial leads for o ∗ n,> ( k ) to a sum of L + 1 floor-functions with decreasing arguments, starting with (cid:22) k − L − L (cid:23) , ending with (cid:22) k − L + L − L (cid:23) = (cid:22) k − L (cid:23) , and corresponding coefficients. In order to find a standard form for this sum we use the followingfloor-identity L − X j =0 (cid:22) k + jL (cid:23) = k, for L ∈ N , and k ∈ Z , (88)which can be seen from L − X j =0 ( k + j ) ( mod L ) = L ( L − k ∈ Z , which is trivial (just add the L termswhich are 0 , , ... , L − mod L and the floor function (cid:22) kL (cid:23) = 1 L ( k − k ( mod L )). This identity allows us to lower L consecutive coefficients of this sum of L + 1 terms by 1, producing a term k if one uses the identity for the first L terms because the second tolast floor-argument is then (cid:4) k +0 L (cid:5) , and the identity is read backwards. If one uses the identity for the last L terms one produces a k − 1, because the last term has (cid:4) k − L (cid:5) . When we apply this identity twice in thedescribed way we pick up n + ( n − 1) = 2 n − ≥ 0, because the coefficientsexcept the first and last one were even and ≥ 2, because these numerator polynomial coefficients resultedfrom first differences of the sequence of odd numbers. Therefore one can extract a factor 2 and if thefloor-functions are written with increasing arguments, starting with (cid:22) kL (cid:23) (the second to last term in theoriginal order), one has exactly the coefficients given by the list F ( n ) of length L of definition 5 . Thisproves the explicit form for o ∗ n,> ( k ). (cid:3) Example 17: O.g.f. G ( x ) G ( x ) = x (1 + 2 ( x + x + x + x + x + x + x ) + x (1 − x ) (1 − x ) = x x (1 − x ) (89)generating, with offset 0, the odd numbers A004273.Before coming to the multiplicative group M odd n we define reduced residue systems M odd n as well asreduced odd residue systems M odd n . See e.g. , [2], p. 113 for the mod n case.33 efinition 6a: Reduced residue system Modd n A reduced residue system M odd n ( RRSn ) is any set of ϕ ( n ) pairwise incongruent M odd n numbers, eachof which is relatively prime to n . E.g. , n = 15, ϕ (15) = 8, { , , , , , , , } or { , , , , , , , } , etc . The first one is thesmallest positive one. These systems will not play a rˆole later on. Definition 6b: Reduced odd residue system Modd n A reduced odd residue system M odd n ( RoddRSn ) is any set of δ ( n ) odd pairwise incongruent M odd n numbers, each of which is relatively prime to n E.g. , n = 15, δ (15) = 4, { , , , } or { , , , } . The first one coincides with M (15) fromeq. (65) and is the smallest positive reduced residue system M odd n . Remember that the parity of themembers in each M odd n residue class, not in class [0], is the same (see lemma 12 ). Later on we willrestrict ourselves mostly to the system M ( n ).We next study the multiplicative group M odd n . The elements are the residue classes [ m j ] , for j =1 , , ..., δ ( n ), corresponding to the members of the reduced residue system M ( n ). In fact, we will takethese representatives, multiplying M odd n , as we have done above.In order to see the group structure one first convinces oneself that this set M ( n ) is closed under M odd n multiplication. This follows from proposition 10 and lemma 12 which showed that these classes haveonly odd numbers, and every odd numbers appear exactly once because of the definition of these classes.The associativity, commutativity and the identity element 1 are also clear. We do not have a formulahow to find the inverse element m − j of m j but coming to this M odd n multiplication from the study ofautomorphisms of the splitting field for the minimal polynomial C ( n, x ) for the algebraic number ρ ( n ),it is clear that these inverses have to exist (the invariance property defines a group). We are dealing herewith the Galois groups for these polynomials. The cycle structure for n = 1 , , ..., 40 is given in table6 . Because each group is Abelian one has for every group of prime order a cyclic group, which can bechecked for the examples given in table 6 . Remark 6: The multiplicative group M odd n is cyclic if δ ( n ) is prime.This is a corollary on the fundamental theorem on finite Abelian groups (see e.g. , [28], p. 49, see also[6], p. 511, Cauchy ’s Theorem A.1.5), or use Lagrange ’s theorem on the order of subgroups, and the factthat the powers of an element, not the identity, generate a cyclic subgroup, to prove that in fact everygroup of prime order is cyclic. The values n for which the order δ ( n ) is prime are given in A215046.This is the sequence [4 , , , , , , , , , , , , , , , , , ... ]. Of course, there areother values n with cyclic M odd n group, like n = 2 , , , , ... . For n = 1 one has the trivial case withcycle structure [[0]], also a cyclic group, viz Z . One can give a more general sequence of n numberswith cyclic G n ∼ = multiplicative M odd n group, namely A210845. Remark 7: The multiplicative group M odd n is cyclic if δ ( n ) is squarefree.The squarefree numbers are given in A005117, namely [1 , , , , , , , , , , , , , , , , , , ... ]. This remark 7 results from the fact that A000688, giving the number of Abelian groups of order n , is 1 exactly for the squarefree numbers A005117. See the formula, based on the H.-E. Richert referencequoted there. Because for each order there is at least the cyclic group these values lead necessarily to acyclic group. The above given A215046 values are a proper subset of those from A210845. There are.however, still more values n with a cyclic M odd n group. Missing are e.g. , 8 , , , , , ... . All the n values with cyclic M odd n group are in A206551. The complementary sequence is A206552, giving the n values with non-cyclic M odd n group. See Table 8 for all values n ≤ n ), the number of groups of order n is 1 if n is a cyclicnumber, in fact, the reverse also holds: if A000001( n ) = 1 then n is a cyclic number. This can be takenas an alternative definition for cyclic numbers because then there is only one (non-isomorphic) group of34his order which has to be the cyclic group. See also Yimin Ge’s Math Blog [YiminGe], where the ‘onlyif’ statement in the proposition may be misleading but in the proof the given statement is correct. E.g. ,for order 6 (not a cyclic number) there are groups other than Z . In fact A000001(6) = 2, and thereis the (non-Abelian) group D(3) (dihedral group). (It is clear that cyclic numbers are not the numbers n for which the multiplicative group mod n , which is Abelian of order ϕ ( n ), is cyclic. These numbersare given in A033948). Because we are dealing with Abelian groups the squarefree numbers are moreinteresting here.Recall the situation for the Galois group for the cyclotomic polynomials (the minimal polynomials for ζ ( n ), an n -th root of unity) which is isomorphic to the multiplicative (Abelian) group of integers modulo n . See Table 7 were we have listed these non-cyclic groups for n ∈ { , ... , } . See also A282624 for aW. Lang link with a table for n ∈ { , ... , } also with generators. This n − value sequence is known asA033949. (The case n = 15 is an exercise in [7], p. 159, Ex. 9.6. 3)a)). The values n with a cyclic Galois group coincide with the moduli n which possess primitive roots r = r ( n ), i.e. , the order of r modulo n is ϕ : r ϕ ( n ) ≡ mod n ), and no smaller positive exponent k satisfies this congruence. These moduli n are known to be exactly p e , p e ′ , e and e ′ . See e.g. , [23], Theorem2.41 , p. 104. All other n lead to non-cyclic multiplicative groups modulo n . See A033949. For thesmallest primitive roots in this case see [1], pp. 864-869, the column called g . We do not know a similarcharacterization of the non-cyclic numbers n for the M odd n group which are shown in Table 8 . However,these numbers have also to appear in A033949. It is clear that we should determine primitive roots r ≡ r ( n ) for the M odd n multiplication, i.e. , find those r from M ( n ) which have order δ ( n ): the smallestpositive k which satisfies r k ≡ +1 ( M odd n ) is k = δ ( n ). See A206550 for these smallest positive primitiveroots M odd n , where a 0 entry, except for n = 1, indicates that there exists no such primitive root. Thissequence starts, with offset 1, as [0 , , , , , , , , , , , , , , , , , , , , , , , , , , ... ]. In generalone does not expect a formula for these primitive roots r = r ( n ), because also in the multiplicationmodulo n case there is no one available. Proposition 14: Number of Modd n primitive roots If a primitive root r for the multiplicative group M odd n exists there are ϕ ( δ ( n )) of them.This is the sequence A216322. For example, n = 13, δ (13) = 6, ϕ (6) = 2 with r = 7 and r = 11; n = 14, δ (14) = 6, ϕ (6) = 2 with r = 5 and r = 11. Proof: If a M odd n primitive root exists then the multiplicative group M odd n is cyclic. The order of thisgroup is δ ( n ), Then the number of pairwise incongruent primitive roots is obtained exactly like in thecase of primitive m-roots of unity on the unit circle by the number of relatively prime numbers less than m which is Euler ’s ϕ ( m ). Here m = δ ( n ). (cid:3) Now the question whether the multiplicative group M odd p with p a prime is cyclic is answered. Up tonow we know from remark 7 the positive answer only for those primes with δ ( p ) = p − 12 squarefree.These are the primes given in A066651. The case p = 2 with δ (2) = 1 is trivially cyclic. For generalprime p one wants to show that there exists a primitive root in the multiplicative M odd p group. This isanalog to the mod p case, where a proof can be found in [23], Theorem 2.36, p. 99, or in [2], ch. 10.4,pp 206 ff. One has, however, to be careful to use only the multiplicative group structure in the proof.Already in connection with lemma 10 we have given a warning to use theorems valid in the modulararithmetic mod n in our case. Another failure occurs for the theorem [23], Theorem 2.3(c) , p. 49, e.g. , x ≡ y ( mod m ) , x ≡ y ( mod m ) if and only if x = y (cid:18) mod m m gcd ( m , m ) (cid:19) . For the M odd n case one hasthe counterexample 19 = 5 ( M odd 3) and 19 = 5 ( M odd 7) but 19 is obviously not congruent 5 ( M odd lemmata , the analoga of mod n facts, are collected before stating proposition 15 . Lemma 19: Analogon of the Fermat-Euler Theorem If a is odd and gcd ( a, n ) = 1 then a δ ( n ) ≡ +1 ( M odd n ), with delta ( n ) = A055034( n ) (see the start of section 3) . 35 roof: Analog to e.g. , [2], Theorem 5.17 , p. 113, with m → n, ϕ → δ , the order of the set (reducedodd residue system) M ( n ) of eq. (65), and modulo → Modd n (we avoid the term Moddulo). Insteadof M ( n ) one can use any other reduced odd residue system M odd n (see definition 6b ), with m j ( n ), j ∈{ , , ... , δ ( n ) } , replaced by any member of the residue class [ m j ( n )]. The cancellation used at the end ofthe proof in [2] can be done here by multiplying with the existing inverses of the b i s there. We are dealingwith the multiplicative group Modd n . (cid:3) Next follows the definition of the order of an element from a reduced odd residue system M odd n . Definition 7: Modd n order of a from a RoddRSn The M odd n order of a positive (odd) integer a from a reduced odd residue system M odd n is the smallestpositive integer h such that a h ≡ M odd n ). E.g. , n = 10 , a = 9 , h = 2 because 9 = 9 ( M odd = 1 ( M odd M odd n orders h of the smallest positive RoddRSn members A216319. Lemma 19 guarantees the existence of a M odd n order h ≤ δ ( n ) for each positive odd number a with gcd ( a, n ) = 1. Example 18: n = 12, δ (12) = 4, M (12) = { , , , } , 1 ≡ M odd ≡ M odd ⇒ ≡ ( M odd n = 12 in table 6 . Lemma 20: The Modd n order divides δ ( n )If h is the M odd n order of a then h | δ ( n ). Moreover, a j ≡ a k ( M odd n ), w.l.o.g. j > k , if and only if h | ( j − k ). Proof: The first part is analog to [23], Corollary 2.32 , p. 98, with ϕ → δ . It uses lemma 19, the Fermat-Euler analogon. Remember that always gcd ( a, n ) = 1 from the order definition. The secondpart, for which one can use the group property, follows from the analog of [23], Lemma 2.31 , p. 98. (Inthe older German version of this book, vol. I, the analog of this lemma 20 appears as Satz 22.3 on p. 63.) (cid:3) Lemma 21: On the Modd p order of a, p a prime If h is the Modd p order of a , with p a prime, one has ( a k ) h ≡ p ) for all k , and 1 = a , a , a , ... , a h − are pairwise incongruent M odd p . Proof: First part: ( a k ) h = ( a h ) k , a h ≡ M odd p ) and M odd p respects multiplication (see proposition10 ). Second part: assume the contrary, i.e. , a i ≡ a j ( M odd p ), 1 ≤ j < i ≤ h − 1. Apply lemma 20 for n = p , showing that h | ( i − j ), but i − j ≤ i ≤ h − (cid:3) Lemma 22: X h ≡ ( Modd p ) has at most h incongruent solutions The congruence X h ≡ M odd p ), h a positive integer, has at most h pairwise incongruent solutions M odd p . Proof: If (cid:22) X h p (cid:23) is even then X h ≡ mod p ). and due to the mod p theorem, e.g. , [21], Theorem42 , p. 80, there are at most h incongruent solutions. If (cid:22) X h p (cid:23) is odd then − X h ≡ mod p ), i.e. , X h ≡ ( p − 1) ( mod p ) which again has at most h incongruent solutions. (cid:3) This is a weak statement, but sufficient for the following. One could try to prove that the number of M odd n incongruent solutions of the congruence X h ≡ M odd p ) is gcd ( h, p − 12 ) if p is odd. Thisnumber is trivially 1 for p = 2. Lemma 23: On the Modd n order of a k If h is the M odd n order of a , then hgcd ( h, k ) is the M odd n order of a k . Proof: This is the analogon of [23] lemma 2.33 , p. 98, using the present lemma 20 , part 2. (cid:3) Now we are ready for the following proposition . 36 roposition 15: Existence of Modd p primitive roots For every prime p there exists a primitive root for the multiplicative group M odd p . Proof: One shows that there are precisely ϕ ( δ ( p )) primitive roots and because this number is always ≥ Theorem 2.36 , p. 99, we infer from the present lemma 19 (with n → p ) thatevery positive odd integer a with gcd ( a, p ) = 1 has a M odd p order h ≡ h p ( a ), 1 ≤ h ≤ δ ( p ) = p − 12 ,and from lemma 20 h | δ ( p ). Moreover, ( a h ) k ≡ M odd p ) for all k , and 1 , a , a ... a h − are h pairwiseincongruent odd numbers M odd p . From lemma 22 these are all the solutions of the congruence X h ≡ M odd p ). Lemma 23 shows that there are ϕ ( h ) numbers a k , with k ∈ { , ... h − } which have M odd p order h ≡ h p ( a ) because ϕ ( h ) is the number of such k with gcd ( k, h ) = 1. The case h = 1 is coveredwith ϕ (1) := 1.With [2], p. 207, we define for each h | δ ( p ) the set A p ( h ) := ( a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a ∈ M ( p ) and M odd p order of a is h ) .There are τ ( δ ( p )) ( τ = A000005, the number of divisors) such disjoint sets. This is the sequence[1 , , , , , , , , , , , , ... ] = A216326. E.g. , p = 7 , M (7) = { , , } , τ (3) = 2 (see alsorow No. n = p = 7 of the M odd n order table A21630). Call ψ p ( h ) := | A p ( h ) | , the number of elementsof this set. We have seen that ψ p ( h ) is 0 or ϕ ( h ) because for each h | δ ( p ) the set A p ( h ) is either empty(no a ∈ M ( p ) has M odd p order h) or this set has ϕ ( h ) elements. Thus ψ p ( h ) ≤ ϕ ( h ). It is clear that X h | δ ( h ) ψ p ( h ) = δ ( p ) because each a ∈ M ( p ) belongs to one of these sets A p ( h ). Now a standard result is X h | δ ( h ) ϕ ( h ) = δ ( p ) (see e.g. , [2], Theorem 2.2 , p. 26, wit h n → δ ( p )), therefore, X h | δ ( h ) ( ϕ ( h ) − ψ p ( h )) = 0,but because ϕ ( h ) − ψ p ( h ) ≥ ψ p ( h ) = ϕ ( h ) > 1, not 0, for each h | δ ( p ). This holdsespecially for h = δ ( p ). Hence the number of primitive M odd p roots is ψ p ( δ ( p )) = ϕ ( δ ( p )) > 1, and theproof is complete. (cid:3) For the sequence δ ( p ( n )) n ≥ , see A130290. For the smallest positive primitive M odd n roots see thesequence A206550. Here for prime n . Corollary 8: Cyclic Modd p group The multiplicative (Abelian) M odd p group, for p a prime, is the cyclic group Z δ ( p ) .We now concentrate on those groups M odd n which are cyclic and notice that whenever δ ( n ) is even thereexists a unique smallest positive odd number > x ≡ M odd n ). This isthe nontrivial solution of this congruence. The trivial one is x = 1 (standing for the class of solutions1 ( M odd n ) which also includes − n . Proposition 16: Nontrivial square-root of 1 (Modd 2 k), k ≥ If n = 2 k , for k ∈ { , , , ... } , and if the Galois group G al ( Q ( ρ ( n )) / Q ) is cyclic, a unique smallestpositive solution r > x ≡ M odd k ) exists, and it is r = n − Proof: For even n = 2 k , k ≥ 2, the cyclic group is then Z δ (2 k ) and δ (2 k ) is even because with thestandard prime factorization δ (2 e p e · · · p e N N ) = 2 e − Y j =1 ..N p e j − j ( p j − N = 0 then e ≥ 2, andif N ≥ e ≥ 1. In both cases δ is even, because in the second case there is at least one j and p j − δ (2 k ) of the cyclic group means that powers of the generator c of the δ -cycle,the smallest positive primitive root (of 1) ( M odd c δ (2 k ) and s := c δ (2 k )2 . This s > x ≡ M odd k ).Because (cid:22) ( n − n (cid:23) = n − n , even, and in order to compute ( n − ( M odd n )37ne has to determine ( n − ( mod n ) , which is 1. Therefore, the unique s for even n is s = n − (cid:3) Example 19: Nontrivial square-root of 1 ( Modd 8 ) s (8) = 7: 7 = 49 ≡ M odd (cid:22) (cid:23) = 6, hence p (49) = +1, and 49 ≡ +1 ( mod ≡ +1 ( M odd n = p , p an odd prime, the cycle length is δ ( p ) = p − 12 . In the event that p − 12 is even, i.e. , if p is of the form 4 k + 1, i.e. , p ≡ mod x ≡ M odd n ) exists. We will numerate the primes congruent 1 ( mod p ( n ) := A002144( n ), n ≥ E.g. , for n = ˆ p (2) = 13 one has, besides the trivial solution x = 1 ( M odd 13) (this residue class includes − x ≡ M odd 13) as a solution. This means that in thegroup table the row for the element 5 has a 1 in the diagonal, like in the row for the identity element 1.These elements are self-inverse like the identity element 1. Note that in the ordinary reduced mod n casethere is always a 1 in the diagonal for row p − 1, besides the 1 in the row for the identity element, but inthat case p − ≡ − mod n ), and hence is a trivial solution. See A206549 for these non-trivial solutionsin our case. The other inverses in these cyclic groups can be read off the cycle structure. One pairs thefirst entry, the generator c of the cycle, with the second to last (the one before the 1, the second withthe third to last, etc. E.g. , for n = 13 one has the inverses 7 − ≡ 11 ( M odd , − ≡ M odd − ≡ M odd 13) , 9 − ≡ M odd s ( p ) of the congruence x ≡ M odd p ) for given prime p = ˆ p ( n ) = A002144( n ) forthe k ( n ) := ˆ p ( n ) − values given in A005098. This k -sequence starts with [1 , , , , , , , , , ... ].The following l -algorithm will determine all these non-trivial solutions s (ˆ p ( n )) of the congruence x ≡ M odd ˆ p ( n )). We start with the Ansatz s ( p ) = p ( o + 1) p − 1, with some odd number o . Thismeans that we are looking for primes of the form s + 1 o + 1 , and we are interested only in primes of theform 4 k + 1, i.e. , ˆ p ( n ). Because s (ˆ p ) = ( o + 1) ˆ p − s (ˆ p ) is odd, say 2 K + 1. This implies2 K ( K + 1) + 1 = o + 12 ˆ p , which shows that o +12 has to be odd, i.e. , o = 4 ˆ l + 1. Now we computethe even number 2 k = p − 12 = (2 K + 1) − (4 ˆ l + 1) ˆ p T ( K ) − l − k (4 ˆ l + 1), wherewe have introduced the triangular numbers T ( K ) := K ( K +1)2 , given in A000217. This implies that2 k (2 ˆ l + 1) = 2 T ( K ) − ˆ l . Therefore, ˆ l is even, say 2 l , and we have, with inputs k, l and K : p = 4 k + 1, s (ˆ p ) = ( o + 1) p − o = 8 l + 1 and s (ˆ p ) = 2 K (ˆ p ) + 1 . Because now2 T ( K ) = 2 k + 2 l (4 k + 1) = 2 l + 2 k (4 l + 1), we find the ‘nice equation’4 T ( K ) + 1 = (4 k + 1) (4 l + 1) . (90)For given prime ˆ p of the form 4 k + 1, i.e. for k = p − (see A005098), we determine the minimal l ∈ N ,such that k (1 + 4 l ) + l produces a triangular number, namely T ( K ) . The index of this triangularnumber is then K = √ T ( K ) + 1 − . We have uses Maple [18] to compute these minimal l values. Itwill be seen from the later proposition 15 that there is always such a minimal l , and not all non-negativevalues appear. Precisely the entries of A094178 and 0 occur as minimal l values. We will from now ondenote this minimal l also by l (hoping to cause no confusion). If l = 0 then k is a triangular number,and conversely. The non-vanishing l values are characterized by having as elements of the squarefreekernel set of 4 l + 1 only distinct primes congruent 1 ( mod E.g. , l = 2 will not appear because thesquarefree kernel of 9 consists of the prime 3 which is not congruent 1 ( mod l values, we use the ‘nice equation’ and observe that38 T ( K ) + 1 = K + ( K + 1) , i.e. , it is the sum to two neighboring squares with gcd ( K, K + 1) = 1(shown by indirect proof, see the remark following definition 3 ). Now a theorem ensures that all primefactors of 4 T ( K ) + 1 are congruent 1 ( mod l + 1 = 1, i.e. , l = 0, or, if l is non-vanishing, then thesquarefree kernel set of 4 l + 1 has only primes congruent 1 ( mod l can be 0 or it is fromA094178, i.e. , from [0 , , , , , , , , , , , , , , , , , , ... ]. The l = 0 value appearsfor the primes [5 , , , , , , , , ... ], which is the sequence A027862. In this case k = ˆ p − is a triangular number. l = 1 for the primes [17 , , , , , , , , , , , ... ] This is thesequence A207337( n ), n ≥ 2. These are the primes ˆ p such that 5 ˆ p = 4 T ( K ) + 1 for some K = K (ˆ p ). E.g. , 5 17 = 85 = 4 21 = 1, thus K (17) = 6. These are also the primes of the form ( m +1) / 10. Just use m = m ( K ) = 2 K + 1. In general, k = T ( K ) − l l + 1 , (see eq. (90) and K = p l + 1) ˆ p + 1 − 12 . E.g. , ˆ p = 53 belongs to l = 1 and K = 11, and T ( K ) = 66. This checks with the ‘nice equation’. Wedo not give here the tables of the sequences l ( n ) = l (ˆ p ( n )) and K ( n ) = K ( l ( n )) (this incorrect renamingof arguments should not lead to confusion) for given ˆ p ( n ) because after the next proposition 17 we willfind another way to obtain the K ( n ) numbers directly, hence the corresponding l ( n ) numbers, and fromthese the desired s (ˆ p ( n )) solutions from s (ˆ p ( n )) = 2 K ( n ) + 1. Proposition 17: Congruence 4 T ( X ) + = X + ( X + ) ≡ ( mod p )There are precisely two incongruent solutions of the congruence f ( X ) := 2 X + 2 X + 1 ≡ mod p ),provided p = ˆ p ( n ) = A002144( n ), n ≥ 1. The smallest positive representative will be called K ( n ), andthe next larger incongruent one is then K n ) := ˆ p ( n ) − − K ( n ). Proof: Because of the degree 2 of f this congruence has at most two incongruent solutions. We shallsee that in fact there are two. This congruence is reduced to a problem of quadratic residues, followinga standard prescription (see, e.g. , [21], pp. 132-3). The discriminant of f is D = 2 − − f by 8 yields (4 X + 2) + 4 ≡ mod p ). With Y = 4 X + 2 this is Y = D ( mod n ),with a composite modulus n = 2 p . This is a quadratic residue problem, but gcd ( D, n ) is not 1, but 4.Theorem 77 of [21] is applied with d = 4, e = 2, f = 1, a + 1 = − n = 2 p Thus the problem isreduced to the quadratic residue problem Z ≡ − mod p ). This is solved by studying the congruencesfor the powers of primes, here just 2 and p , separately (see e.g. , [21], sect. 26, pp. 83-5). The congruencemodulo 2 has only the solution +1 (because − ≡ +1 ( mod p one can consult the Legendre symbol (cid:18) − p (cid:19) which is ( − p − (see e.g. , [23], Theorem . − p if and only if this symbol is +1, demanding that p ≡ mod i.e. , p = ˆ p from A002144. Call the smallest positive solution x , then ˆ p − x is also an incongruent solution moduloˆ p , and two is the maximal number of solutions because of the degree 2 of this congruence. This impliesthat there are 1 · p (see e.g. , [21], Theorem 46,p. 84). Returning to the original problem this proves that there are also two incongruent solutions. If thesmallest positive solution for ˆ p ( n ) = A207337( n ) is called K ( n ), then the next larger incongruent solutionof 4 T ( X ) + 1 = X + ( X + 1) ) ≡ mod ˆ p ( n )) is K n ) := ˆ p ( n ) − − K ( n ), which is obvious. (cid:3) The pair of sequences of all positive solutions modulo 5 , , and 17 are given in A047219, A212160 andA212161, respectively, where in each case the even indexed members are the positive solutions congruentto K ( n ) and the odd indexed ones are the positive solutions congruent to K n ), i.e. , a (2 k ) = k ˆ p ( n ) + K ( n ) and a (2 k + 1) = k ˆ p ( n ) + K n ), k ≥ 0. For the three given examples n = 1 , , K ( n ) , K n )) have been computed with Maple [18] and they can be found as A212353and A212354. The first entries are for K : [1 , , , , , , , , , , , , , , , , , ... ] and forK2: [3 , , , , , , , , , , , , , , , , , , ... ] corresponding to the primes ˆ p :[5 , , , , , , , , , , , , , , , , , , ... ]. Note that proposition 17 yieldsdirectly the searched for nontrivial solution of s (ˆ p ( n )) ≡ +1 ( M odd ˆ p ( n )) via s (ˆ p ( n )) = 2 K ( n ) + 1.Compare this with the s values given in A206549 with first entries [3 , , , , , , , , , , , , , , , , , , ... ].39ecause ˆ s ( n ) := s (ˆ p ( n )) = 2 K ( n ) + 1 ≤ ˆ p ( n ) − i.e. , K ( n ) ≤ ˆ p − 32 iff b s n ) := K n ) + 1 ≥ ˆ p ( n )and ˆ s ( n ) ≤ b s n ), n ≥ 1. Thus only ˆ s ( n ) ∈ M (ˆ p ), the restricted odd residue class M odd ˆ p , and thesolution b s n ) is discarded.This proposition and the ‘nice equation’ eq. (90) show that the l − algorithm from above will indeed pro-duce a solution l , related to the existing K ( n ) for given ˆ p ( n ), via l ( n ) = (cid:18) T ( K ( n )) + 1ˆ p ( n ) − (cid:19) / 4. Thesmallest positive non-trivial solution s (ˆ p ( n )) of the congruence x ≡ +1 ( M odd ˆ p ( n )) is then 2 K ( n ) + 1.See K ( n ) = A212353( n ), n ≥ E.g. , n = 5 , ˆ p (5) = 37 , K (5) = 15, l = (cid:22) , + 137 (cid:23) − / s ( n ) = 31, with (cid:22) (cid:23) = 25 = o and 31 = 961 = − mod 37) = 36.As a application involving s (ˆ p ( n )) we derive the analog of Wilson ’s theorem Q R ( p ) = ( p − ≡ ( p − 1) ( mod p ) ≡ − mod p ), for each prime p (see e.g. , [2], Theorem 5.24, p. 116, or [10], Theorem 80,p. 68), where R ( n )is the set of the representatives of the smallest positive reduced residue system mod n for n ≥ ϕ ( n ). In the M odd p case we have to replace the set R ( n ) by the set M ( n ) ofeq. (65) with order δ ( n ). Proposition 18: Analog of Wilson’s theorem for Modd p Y M ( p ) = ( p − ≡ p − is oddˆ s ( n ) if p − is even, p = ˆ p ( n ) , n ∈ N ( M odd p ) , (91)where ˆ s ( n ) = s (ˆ p ( n )), ˆ p ( n ) = A002144( n ), stands for the above treated nontrivial root M odd n , i.e. , thesolution of the congruence x ≡ M odd n ) which is not 1, if it exists. Note that − ≡ +1 ( M odd n ), n ≥ lemma 10 ). For the double factorials ( p − Proof : The multiplicative group group M odd p , p a prime, is the cyclic group Z δ ( p ) from proposition14 . The order δ ( p ) of this group is even if and only if p = ˆ p , i.e. , a prime 1 ( mod s ( n ) = s (ˆ p ( n )) exists, and an algorithm for finding it has beengiven. Besides the unit element 1 and this ˆ s ( n ) all other factors in Q M (ˆ p ) can be paired such thattheir product is 1 ( M odd ˆ p ) (see the discussion after proposition 16 ), leaving only s (ˆ p ). In the other case,when p ≡ mod p ), the group order is odd. Then all numbers besides 1 can be paired in the product toproduce 1 ( M odd p ), and the result is therefore 1. (cid:3) For the cyclic Galois groups belonging to C ( n, x ) the list of the smallest positive primitive roots r ( n )( i.e. , the smallest element from M ( n ) which generate these cyclic groups) are found under A206550. Asmentioned above, we do not have a formula for those n > n = [6 , , , , , , , , , , , , , ... ]. Open problems: • Proof of the conjecture on the q − sequence related to the discriminant of the minimal C ( n, x ) polyno-mials; or find a counterexample. • Characterization of the values n for which the Galois group G n is non-cyclic. • Characterization of the values n for which the cycle of the cyclic multiplicative group M odd n ∼ = G n isgenerated by 3. • More theorems on multiplicative M odd n arithmetic.40 ppendix A The proof of eq. (30), with ϕ ( n ) n replaced by the product-formula given there a bit later, is a (nice)application of PIE (the principle of inclusion and exclusion. See e.g. , [5], Theorem 4.2, pp. 134 ff).As mentioned above, this formula uses only the distinct prime factors of n , the elements of the set sqf kset ( n ), the set of primes of the squarefree kernel of n . The product formula for ϕ ( n ) n can be readas the generating function for the elementary symmetric functions (here polynomials) in the variables p j , j = 1 , , ..., M ( n ). M ( n ), also called ω ( n ), is given in [27] as A001221( n ). ϕ ( n ) n = M ( n ) Y j =1 (cid:18) − p j (cid:19) = M ( n ) X r =0 ( − r σ r (cid:18) p , ..., p M ( n ) (cid:19) , (92)with σ = 1, and symbolically σ r = X r p . · · · p . , where the sum extends over the (cid:18) Mr (cid:19) terms with r factors p . with increasing indices. E.g. , M = 3 with σ = p p + p p + p p .To calculate n − X k =1 gcd ( k,n )=1 k one starts, at the zeroth step ( r = 0), with the unrestricted sum which is n ( n − n which are ≤ ( n − p j -multiples this sum is p j npj − X k =1 k = n (cid:18) np j − (cid:19) , for j = 1 , , ..., M ( n ). Thisleads, in step r = 1 to the subtraction of n X j (cid:18) np j − (cid:19) . Now in this subtraction all multiples of theproduct of two different p j s appeared twice, therefore one has, in the next step ( r = 2), to add themonce. This is done by +1 X i < j p i p j npi pj − X k =1 k = n X i < j (cid:18) np . p . − (cid:19) . In step r = 3 one concentrateson products of three different primes p i p i p i with i < i < i . Now such a 3 − product appearedonce in step r = 0 ( in the unrestricted sum), − r = 1, originating from the multiples( p i p i ) · p i , ( p i p i ) · p i and ( p i p i ) · p i , and +3 times in step r = 2, from the multiples p i · ( p i p i ), p i · ( p i p i ), and p i · ( p i p i ). Therefore, up to this stage, each such 3 − product appeared once toomuch, and it is subtracted in this step r = 3 when all these 3 − products terms are summed. Now thepattern starts to become clear. Up to, and including, step r = 3 one has for each 4 − product the counting1 − − − 1. Hence in step r = 4 one adds once the sum over all multiples of each such4 − product. In general, in step r this counting will produce ( − r +1 (from the alternating row r = 4 inthe Pascal triangle A007318), and one will therefore add ( − r times the sum over all multiples of eachsuch r − product. This yields n − X k =1 gcd ( k,n )=1 k = n " ( n − − X i (cid:18) np i − (cid:19) + X i
For the proof of eq. (56) we follow [24], p.33, and consider the rewritten version n − Y k =1 (cid:16) − e π i kn (cid:17) = n is odd ,0 if n is even . (96)In order to see that this is identical to eq. (56), just extract e π i kn , leading to the (2 cos) factors underthe product, and in front summing in the exponent leads to the factor e π i ( n − = i n − = ( − n − .The l.h.s. of eq. (96) is then computed using (for the last step see e.g. , [9], Exercise 50, p. 149)1 + z + z + ... + z n − = z n − z − n − Y k =1 ( z − e π i kn ) , (97)where one specializes to z = − 1. (By putting z = +1 one finds the result n − Y k =1 (cid:18) π kn (cid:19) = n , n ≥ n = 1.) 42 eferences [1] Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions , National Bureau of Stan-dards Applied Math. Series 55, Tenth Printing, reprinted as Dover publication, seventh printing1968, New York, 1972[2] T. Apostol, Introduction to Analytic Number Theory , Springer, 1986[3] E. Artin, with a section by N. A, Milgram, Galoissche Theorie, 3. Auflage, Harri Deutsch, 1988.English version: Galois theory. Lectures delivered at the University of Notre Dame, 1966[4] Chan-Lye Lee and K. B. Wong, On Chebyshev’s polynomials and certain combinatorial identities,2009, Bull. Malaysian Sciences Soc. .[5] Ch. A. Charalambides, Enumerative Combinatorics , Chapman &Hall/CRC, 2002[6] D. A. Cox, Galois Theory , Wiley, 2004[7] J.-P. Escofier, Galois Theory , Springer, 2001[8] Yimin Ge’s Math Blog, 2009, http://yiminge.wordpress.com/2009/01/22/all-groups-of-order-n-are-cyclic-iff/ [9] R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics , Addison-Weseley, Reading,Massachusetts, 1991[10] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, fifth ed,. Oxford Sciencepublications, 2003[11] The On-Line Encyclopedia of Integer Sequences T M (OEIS), published electronically at http://oeis.org , 2011[12] S. Lang, Algebra, revised third ed., Springer, 2002[13] W. Lang, Minimal Polynomials for cos (cid:18) πn (cid:19) , a link under [27] A181875.[14] W. Lang, Divisor Product Representation for Natural Numbers, a link under [27] A007955.[15] W. Lang, A181878, Coefficient array for square of Chebyshev S-polynomials, a link under[27] A181878.[16] A. Lazarev, Galois Theory, lecture notes with problems and solutions, University of Leicester, [17] D. H. Lehmer, A Note on Trigonometric Algebraic Numbers, Am. Math. Monthly 40,8 (1933) 165-6.[18] Maple T M , [19] Wolfram Mathworld, http://mathworld.wolfram.com/SylvesterCyclotomicNumber.html ,Sylvester Cyclotomic Number.[20] W. Magnus, F. Oberhettinger, R. P. Soni, Formulas and theorems for the special functions of math-ematical physics. 3 rd enlarged ed., 1966, Springer, Berlin.[21] T. Nagell, Introduction to Number Theory , Chelsea Pub. Comp., N.Y., 2nd ed., 1964.4322] I. Niven, Irrational Numbers , The Math. Assoc. of America, second printing, distributed by JohnWiley and Sons, 1963.[23] I. Niven, H. S. Zuckerman and H. L. Montgomery, An Introduction to the Theory Of Numbers , FifthEdition, John Wiley and Sons, Inc., NY, 1991.[24] R. Remmert, Funktionentheorie 2 , Springer, 1991[25] P. Ribenboim, Classical Theory of Algebraic Numbers , Springer, 2001.[26] Th. J. Rivlin, Chebyshev Polynomials. From Approximation Theory to Algebra and Number Theory ,second edition, Wiley-Interscience, 1990.[27] The On-Line Encyclopedia of Integer Sequences (2010), published electronically at http://oeis.org .[28] A. Speiser, Die Theorie der Gruppen endlicher Ordnung , Vierte Auflage, 1956, Birkh¨auser, Basel[29] P. Steinbach, Golden Fields: A Case for the Heptagon, Mathematics Magazine, 70,1 (1997) 22-31, .[30] W. Watkins and J. Zeitlin, The Minimal Polynomial of cos(2 π/n ), Am. Math. Monthly 100,5 (1993)471-4, .[31] Wikipedia, Cycle graph (algebra) http://en.wikipedia.org/wiki/Cycle_graph_(algebra) and List of small groups http://en.wikipedia.org/wiki/List_of_small_groups .Keywords: regular n -gons, algebraic number, minimal polynomial, zeros, factorization, congruences, Galois theory, cycle graphs.AMS MSC numbers: 11R32, 11R04, 08B10, 13F20, 12D10, 13P05OEIS A-numbers: A000001, A000005, A000007, A000010, A000012, A000035, A000265, A000668, A000688,A001221, A002110, A002144, A003277, A004124, A005013, A005098, A005117, A006053, A006054,A007310, A007318, A007775, A007814, A007947, A007955, A008683, A023022, A024556, A027862,A033949, A045572, A049310, A0503384, A052547, A047219, A053120, A055034, A066651, A076512,A077998, A084101, A085810, A088520, A090298, A092260, A094178, A106803, A109395, A110551,A113807, A116423, A120757, A127672, A128672, A130290, A130777, A147600, A162699, A181875,A181876, A181878, A181879, A181880, A193376, A193377, A193679, A193680, A193681, A193682,A203571, A203572, A203575, A204453, A204454, A204457, A204458, A206543, A206544, A206545,A206546, A206547, A206548, A206549, A206550, A206551, A206552, A207333, A207334, A207337,A210845, A212160, A212161, A212353, A212354, A215041, A215046, A216319, A216320, A216322,A216326, A255237. A282624. 44 able 1: Reduced DSR-algebras (over Q ), n = , ..., n ρ ≡ ρ ( n ) reduced DSR-algebra δ ( n ) DSR − basis3 ρ = 1 1 < > √ ρ = 2 2 < , ρ > ϕ = 12 (1 + √ ρ = ρ + 1 2 < , ρ > √ ρ = 3 , [ σ = ρ − < , ρ > (cid:0) π (cid:1) ρ = 1 + σ, σ = 1 + ρ + σ, ρ σ = ρ + σ < , ρ, σ > p √ ρ = 1 + σ, σ = 2 σ + 1 , τ = 2 (1 + σ ) , < , ρ, σ, τ >ρ σ = ρ + τ, ρ τ = 2 σ, σ τ = 2 ρ + τ ,( σ = 1 + √ , τ = √ ρ ) (cid:0) π (cid:1) ρ = 1 + σ, σ = 2 + ρ + σ, ρ σ = 2 ρ + 1 , < , ρ, σ > ( σ = ρ − , [ τ = ρ ( σ − 1) = 1 + ρ ] ϕ √ − ϕ ρ = 1 + σ, σ = − σ, τ = − σ, < , ρ, σ, τ >ρ σ = ρ + τ, ρ τ = 3 σ − , σ τ = ρ + 2 τ ,( σ = ρ − ϕ, τ = ρ ( σ − ϕ ) √ − ϕ )[ ω = 2 ( − σ ) = 2 ϕ ] (cid:0) π (cid:1) ρ = 1 + σ, σ = 1 + σ + ω, < , ρ, σ, τ, ω >τ = 1 + σ + τ + ω, ω = 1 + ρ + σ + τ + ω , ρ σ = ρ + τ, ρ τ = σ + ω, ρ ω = τ + ω , σ τ = ρ + τ + ω, σ ω = σ + τ + ω , τ ω = ρ + σ + τ + ω ,( σ = ρ − , τ = ρ ( σ − , ω = σ ( σ − − p √ ρ = 1 + σ, σ = 2 (1 + τ ) , τ = 3 (1 + σ ) , < , ρ, σ, τ >ρ σ = ρ + τ, ρ τ = 1 + 2 σ, σ τ = 3 ρ + τ ,[ ω = 1 + σ, χ = 2 ρ ],( σ = ρ − √ τ = √ 22 (3 + √ ρ ( n ) := 2 cos (cid:16) πn (cid:17) , R ( n ) k = S ( k − , ρ ( n )) , k = 1 , ..., j n k ,R = 1 , R ≡ ρ, R ≡ σ, R ≡ τ, R ≡ ω, R ≡ χ (dependence on n suppressed).In round brackets the values for the basis elements are given in terms of ρ . In square brackets the lineardependent DSRs are given. Boxed n -numbers indicate linear dependent DSRs.45 able 2: Minimal polynomials of 2 cos (cid:16) π n (cid:17) for n = , , ..., . n C ( n , x ) x + 2 x x − x − x − x − x − x − x − x + 1 x − x + 2 x − x − x − x + 5 x − x − x + 3 x + 3 x − x − x + 1 x − x − x + 4 x + 6 x − x − x − x + 14 x − x + x − x − x + 1 x − x + 20 x − x + 2 x − x − x + 6 x + 15 x − x − x + 4 x + 1 x − x + 9 x − x − x − x + 7 x + 21 x − x − x + 10 x + 5 x − x − x + 19 x − x + 1 x + x − x − x + 8 x + 8 x + 1 x − x + 44 x − x + 55 x − x − x − x + 9 x + 36 x − x − x + 35 x +35 x − x − x + 1 , x − x + 20 x − x + 1 x − x + 35 x − x − x + 5 x + 25 x − x − x − x + 65 x − x + 182 x − x + 13 x − x + 27 x − x + 9 x − x − x + 53 x − x + 86 x − x + 1 x − x − x + 12 x + 66 x − x − x +120 x + 210 x − x − x + 56 x + 28 x − x − x − x + 14 x − x + 1... 46 able 3: A187360 ( n , m ) coefficient array ofminimal polynomials of 2 cos (cid:16) π n (cid:17) , rising powers n/m 0 1 2 3 4 5 6 ... -1 1 -2 0 1 -1 -1 1 -3 0 1 -1 -3 0 1 -1 3 3 -4 -1 1 -1 -3 6 4 -5 -1 1 -7 0 14 0 -7 0 1 able 4: Zeros of C ( n , x ) in power basis (rising powers of ρ ( n ) ) for n = , , ..., n coefficients of C-zeros in power basis < ρ , ..., ρ δ ( n ) − > [[ − [[0]] [[1]] [[0 , , [0 , − [[0 , , [1 , − [[0 , , [0 , − [[0 , , [ − , − , , [2 , , − [[0 , , [0 , − , , , [0 , , , − , [0 , − [[0 , , [ − , − , , [2 , , − [[0 , , [0 , − , , , [0 , , , − , [0 , − [[0 , , [0 , − , , , [1 , , − , − , , [ − , , , , − , [2 , , − [[0 , , [0 , , , − , [0 , − , , , [0 , − [[0 , , [0 , − , , , [0 , , , − , , , [1 , − , − , , , − , [ − , , , , − , [2 , , − [[0 , , [0 , − , , , [0 , , , − , , , [0 , − , , , , − , [0 , , , − , [0 , − [[0 , , [ − , , , − , [ − , − , , , [2 , , − [[0 , , [0 , − , , , [0 , , , − , , , [0 , − , , , , − , , , [0 , , , − , , , , − , [0 , − , , , , − , [0 , , , − , [0 , − [[0 , , [0 , − , , , [0 , , , − , , , [0 , − , , , , − , , , [ − , , , − , − , , , − , [2 , , − , , , , − , [ − , , , , − , [2 , , − [[0 , , [0 , , , − , , , [0 , − , , , , − , [0 , , , − , , , [0 , − , , , , − , [0 , − [[0 , , [0 , − , , , [0 , , , − , , , [0 , − , , , , − , , , [1 , , − , − , , , − , − , , [ − , , , , − , , , , − , [2 , , − , , , , − , [ − , , , , − , [2 , , − [[0 , , [0 , − , , , [0 , − , , , , − , , , [0 , , , − , , , , − , [0 , − , , , , − , , , [0 , , , − , , , , − , [0 , , , − , [0 , − [[0 , , [0 , , , − , , , [2 , , − , − , , [ − , − , , , , − , [ − , , , , − , [2 , , − [[0 , , [0 , − , , , [0 , , , − , , , [0 , − , , , , − , , , [0 , , , − , , , , − , , , [0 , − , , , , − , , , , − , [0 , , , − , , , , − , [0 , − , , , , − , [0 , , , − , [0 , − [[0 , , [0 , − , , , [0 , , , − , , , [0 , − , , , , − , , , [0 , , , − , , , , − , , , [ − , − , , , − , − , , , − , − , , [2 , , − , , , , − , , , , − , [ − , , , , − , , , , − , [2 , , − , , , , − , [ − , , , , − , [2 , , − [[0 , , [0 , , , − , , , [0 , − , , , , − , , , [0 , − , , , , − , [0 , , , − , , , [0 , , , − , , , , − , [0 , − , , , , − , [0 , − [[0 , , [0 , − , , , [0 , − , , , , − , , , [0 , , , − , , , , − , , , [0 , − , , , − , − , , , , − , [0 , , − , − , , , − , − , , [ − , , , , − , , , , − , [2 , , − , , , , − , [ − , , , , − , [2 , , − [[0 , , [0 , − , , , [0 , , , − , , , [0 , − , , , , − , , , [0 , , , − , , , , − , , , [0 , − , , , , − , , , , − , , , [0 , , , − , , , , − , , , , − , [0 , − , , , , − , , , , − , [0 , , , − , , , , − , [0 , − , , , , − , [0 , , , − , [0 , − [[0 , , [0 , , , − , , , [0 , − , , , , − , , , [ − , , , − , , , , − , [2 , − , − , , , − , [2 , − , − , , , , − , , , [ − , , , , − , , , , − , [ − , , , , − , [2 , , − continued on next pageNote: the higher power coefficients not shown are all zero . Example: n = 27: the first zero of C ( , x ) is ρ (27) , the second zero is 5 ρ ( ) − ρ ( ) − ρ ( ) , etc. able 4 continued: Zeros of C ( n , x ) in power basis (rising powers of ρ ( n ) ) forn = , ..., n power basis coefficients for < ρ , ..., ρ δ ( n ) − > [[0 , , [0 , − , , , [0 , , , − , , , [0 , , , − , , , , − , , , [0 , − , , , , − , , , , − , , , [0 , , , − , , , , − , , , , − , [0 , − , , , , − , , , , − , , , [0 , , , − , , , , − , , , , − , [0 , − , , , , − , , , , − , [0 , − , , , , − , [0 , , , − , [0 , − [[0 , , [0 , − , , , [0 , , , − , , , [0 , − , , , , − , , , [0 , , , − , , , , − , , , [0 , − , , , , − , , , , − , , , [0 , , , − , , , , − , , , , − , , , [1 , − , − , , , − , − , , , − , − , , , − , [ − , , , , − , , , , − , , , , − , [2 , , − , , , , − , , , , − , [ − , , , , − , , , , − , [2 , , − , , , , − , [ − , , , , − , [2 , , − [[0 , , [0 , − , , , , − , , , [0 , − , , , , − , , , [0 , , , − , , , , − , [0 , − , , , , − , , , [0 , , , − , , , , − , [0 , , , − , , , , − , [0 , − able 5: Extended set c M ( n ) , first differences △ c M ( n ) , and floor-pattern F ( n ) for composed odd squarefreemodulus n. m n ( m ) δ (n) c M ( n ) △ c M ( n ) F ( n ) , , , , , 17] [1 , , , , 4] [2 , , , , , , , 12 [0 , , , , , , , 23] [1 , , , , , , 4] [1 , , , , , , , , , , , 20 [0 , , , , , , , , , , , 35] [1 , , , , , , , , , , 4] [1 , , , , , , , , , , , , , , , , , , , 24 [0 , , , , , , , , , , , , , 37] [1 , , , , , , , , , , , , 4] [0 , , , , , , , , , , , , · , , , , , , , , , , , 24 [0 , , , , , , , , , , , , , 41] [1 , , , , , , , , , , , , 4] [1 , , , , , , , , , , , , · 13 0 , , , , , , , , , , , 32 [0 , , , , , , , , , , [1 , , , , , , , , , , , , , , , , 4] [1 , , , , , , , , , , , , , , , , · 17 31 , , , , , , , 53] 0 , , , , , , , , , , , , , , , 40 [0 , , , , , , , , , , , , [1 , , , , , , , , , , , [0 , , , , , , , , , , , , , , · 11 29 , , , , , , , , , , 57] 2 , , , , , , , , , 4] 0 , , , , , , , , , , , , , , , , , , , , , , , , , 36 [0 , , , , , , , , , , , [1 , , , , , , , , , [1 , , , , , , , , , , , , , , , · 19 31 , , , , , , , , , 59] 2 , , , , , , , , , 4] 1 , , , , , , , , , , , , , , , , , , , , 48 [0 , , , , , , , , , , , , , , [1 , , , , , , , , , , , , [0 , , , , , , , , , , , , , , , , · 13 33 , , , , , , , , , , , , 67] 2 , , , , , , , , , , , , 4] 0 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 10 69 44 [0 , , , , , , , , , , , , , [1 , , , , , , , , , , , , [1 , , , , , , , , , , , , , , , · 23 41 , , , , , , , , , , 71] 4 , , , , , , , , , , 4] 1 , , , , , , , , , , , , , , , , , , , , , , , , , , , , 11 77 60 [0 , , , , , , , , , , , , [1 , , , , , , , , , , , , , , , , [0 , , , , , , , , , , , , , , , · 11 29 , , , , , , , , , , , , , , , , , , , , , , , , 4] 0 , , , , , , , , , , , , , , , , , , , , , , , , 79] 0 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , continued able 5 ctnd.: Extended set c M ( n ) , first differences △ c M ( n ) , and floor-pattern F ( n ) for composed odd squarefreemodulus n. m n ( m ) δ (n) c M ( n ) △ c M ( n ) F ( n ) 12 85 64 [0 , , , , , , , , , , , , , [1 , , , , , , , , , , , [0 , , , , , , , , , , , , , , , , · 17 33 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 87] 2 , , , , , , , , , , 4] 0 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 13 87 56 [0 , , , , , , , , , , 31 [1 , , , , , , , , , , , , , , , [1 , , , , , , , , , , , , , , · 29 35 , , , , , , , , , , , , , , , , , , , , , , , 4] 0 , , , , , , , , , , , , , , , , , , , , , 89] 0 , , , , , , , , , , , , , , , , , , , , , , , , , , , 14 91 72 [0 , , , , , , , , , , , [1 , , , , , , , , , , , , , [0 , , , , , , , , , , , , , , , · 13 27 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 4] 0 , , , , , , , , , , , , , , , , , , , , , , , 93] 0 , , , , , , , , , , , 15 93 60 [0 , , , , , , , , , , , , [1 , , , , , , , , , , , [1 , , , , , , , , , , , , , , , · 31 37 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 95] 4 , , , , , , , , 4] 0 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 16 95 72 [0 , , , , , , , , , , , [1 , , , , , , , , , , , , , [0 , , , , , , , , , , , , , , , · 19 29 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 4] 0 , , , , , , , , , , , , , , , , , , , , , , , 97] 0 , , , , , , , , , , , , , , , , , , , , , , , , , , 17 105 48 [0 , , , , , , , , , , [1 , , , , , , , , , [4 , , , , , , , , , , , , · · , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 4] 0 , , , , , , , , , , , , , , , , , , , , , , , able 6: Cycle structure of G al ( Q ( ρ ( n )) / Q ) for n = , , ..., n cycles1 [[0]] = [[1]] [[1]] [[1]] [[ , [[ , [[ , [[3 , , [[3 , , , [[5 , , [[3 , , , [[3 , , , , [[ , , [ , , [ , [[7 , , , , , [[5 , , , , , [[7 , , , [[3 , , , , , , , [[3 , , , , , , , [[5 , , , , , [[3 , , , , , , , , [[3 , , , , [7 , , , , [ , , [ , [[11 , , , , , [[3 , , , , , , , , , [[3 , , , , , , , , , , [[5 , , , , [ , , [11 , , , , [ , [[3 , , , , , , , , , [[7 , , , , , , , , , , , [[5 , , , , , , , , [[3 , , , , , , [5 , , , , , , [17 , , , , , [[3 , , , , , , , , , , , , , [[7 , , , , [13 , , , , [ , , [ , [[3 , , , , , , , , , , , , , , [[3 , , , , , , , , , , , , , , , [[5 , , , , , , , , , [[3 , , , , , , , , , , , , , , , [[3 , , , , , , , , , , , [[5 , , , , , , [7 , , , , , , [11 , , , , , [[5 , , , , , , , , , , , , , , , , , [[13 , , , , , , , , , , , , , , , , , [[7 , , , , , , , , , , , [[3 , , , , [7 , , , , [11 , , , , [13 , , , , [17 , , , , [19 , , , n -numbers indicate non-cyclic Galois groups. See Table 8.Boldface numbers are nontrivial square roots Modd n , denoted by s in the text.52 able 7: Non-cyclic Galois groups G al ( Q ( ζ ( n )) / Q ) , n ≤ n ϕ ( n ) cycle structure no. of cycles Galois group8 Z × Z Z × Z Z × Z Z × Z Z × Z 12 6 Z × Z Z 12 6 Z × Z Z × Z 16 8 Z × Z 20 10 Z × Z 24 12 Z × Z × Z 12 6 Z × Z 24 12 Z × Z × Z 16 4 Z × Z 12 6 Z × Z 20 10 Z × Z 24 12 Z × Z × Z 16 4 Z × Z 32 16 Z × Z 24 12 Z × Z × Z 40 20 Z × Z × Z 24 6 Z × Z 36 18 Z × Z 16 4 Z × Z 36 6 Z × Z 32 16 Z × Z 48 12 Z × Z 20 10 Z × Z 32 16 Z × Z 44 22 Z × Z 24 12 Z × Z × Z 24 6 Z × Z 40 20 Z × Z × Z 36 18 Z × Z 60 30 Z × Z × Z 24 12 Z × Z × Z 32 4 Z × Z Continued on the next page. 53 able 7 continued: Non-cyclic Galois groups G al ( Q ( ζ ( n )) / Q ) , n ≤ n ϕ ( n ) cycle structure no. of cycles Galoisgroup84 24 6 Z × Z 64 16 Z × Z 56 28 Z × Z × Z 40 10 Z × Z 24 12 Z × Z × Z 72 12 Z × Z × Z 44 22 Z × Z 60 30 Z × Z × Z 72 36 Z × Z × Z 32 8 Z × Z 60 30 Z × Z × Z 40 20 Z × Z × Z ...The cyclic group of order m is denoted by Z m . For all other values n ≤ the Galois group is thecyclic group Z ϕ ( n ) .Only independent cycles are counted, i.e., cycles which appear as sub-cycles of the given ones have beenomitted.The notation, e.g., , means that there are cycles of order (length) , one cycle of order ,one cycle of order and two cycles of order .Direct products of identical cyclic groups are sometimes written in exponent form, e.g., Z stands for Z × Z .Boxed and colored n -numbers indicate where some non-cyclic Galois group appears for the first time.Some of the cycle graphs are shown in Fig. 4. 54 able 8: Non-cyclic Galois groups G al ( Q ( ρ ( n )) / Q ) , n ≤ n δ ( n ) cycle structure no. of cycles Galoisgroup12 Z × Z Z × Z Z × Z 12 6 Z × Z Z × Z 12 6 Z × Z 16 4 Z × Z 12 6 Z × Z 20 10 Z × Z 16 8 Z × Z 24 12 Z × Z × Z 24 12 Z × Z × Z 16 4 Z × Z 18 6 Z × Z 24 12 Z × Z × Z 20 10 Z × Z 32 16 Z × Z 24 12 Z × Z × Z 24 12 Z × Z × Z 36 18 Z × Z 24 12 Z × Z × Z 32 8 Z × Z 24 6 Z × Z 32 16 Z × Z 40 20 Z × Z × Z 24 12 Z × Z × Z 36 12 Z × Z 44 22 Z × Z 32 16 Z × Z 40 20 Z × Z × Z ...The cyclic group of order m is denoted by Z m . For all other n ≤ cases the Galois group is thecyclic group Z δ ( n ) . The n values are given in A206552.Only independent cycles are counted, i.e., cycles which appear as sub-cycles of the given ones have beenomitted.The notation, e.g., , means that there are cycles of order (length) , one cycle of order ,one cycle of order and two cycles of order .Direct products of identical cyclic groups are sometimes written in exponent form, e.g., Z stands for Z × Z .Boxed and colored n -numbers indicate where some non-cyclic Galois group appears for the first time.For the cycle graphs see Fig. 4. 55 =8: Z_4 x Z_2 δ =4: Z_2 x Z_2 δ =12: Z_3 x Z_2 x Z_2 δ =16: Z_4 x Z_4 δ =20: Z_5 x Z_2 x Z_2 δ =16: Z_8 x Z_2 δ =24: Z_4 x Z_3 x Z_2 δ =16: Z_4 x Z_2 x Z_2 δ =18: Z_3 x Z_3 x Z_2 δ =32: Z_16 x Z_2 δ =36: Z_9 x Z_2 x Z_2 Figure 4: Cycle graphs for non-cyclic Galois groups G al ( Q ( ρ ( n )) / Q ) appearing for n = .. . δ is thedegree of C ( n , x ), the maximal polynomial of ρ ( n ), hence the order of the Galois group. See Table 8.Continued on next page. 56 =32: Z_8 x Z_4 δ=24: Z_3 x Z_2 x Z_2 x Z_2 δ=40: Z_5 x Z_4 x Z_2δ=36: Z_4 x Z_3 x Z_3 Figure 4 continued: Cycle graphs for non-cyclic Galois groups G al ( Q ( ρ ( n )) / Q ) appearing for n = ..100