aa r X i v : . [ m a t h . N T ] J u l A PROOF OF THE LANDSBERG-SCHAAR RELATIONBY FINITE METHODS
BEN MOORE
Abstract.
The Landsberg-Schaar relation is a classical identitybetween quadratic Gauss sums, normally used as a stepping stoneto prove quadratic reciprocity. The Landsberg-Schaar relation it-self is usually proved by carefully taking a limit in the functionalequation for Jacobi’s theta function. In this article we present adirect proof, avoiding any analysis. Introduction
The aim of this article is to prove, using only techniques of elemen-tary number theory, the Landsberg-Schaar relation for positive integral a and b :1 √ a a − X n =0 exp (cid:18) πin ba (cid:19) = 1 √ b exp (cid:18) πi (cid:19) b − X n =0 exp (cid:18) − πin a b (cid:19) . This relation was first discovered in 1850 by Mathias Schaar [11], whoproved it using the Poisson summation formula, and proceeded to de-rive from it the law of quadratic reciprocity. In 1893 Georg Landsberg,apparently unaware of Schaar’s work, rediscovered a slightly more gen-eral version of the relation [8]. Although Landsberg emphasises the roleof modular transformations, his proof is closer in spirit to the modernone given in [9], in which one takes a limit of the functional equationfor Jacobi’s theta function towards rational points on the real line.A few remarks are in order concerning some closely related resultsinvolving techniques differing from those in the present article. Firstly,whilst this article was under review, the author noticed that in thearticle [2], the authors prove Hecke’s generalisation of the Landsberg-Schaar identity over number fields. Their argument is elementary ex-cept for an appeal to Milgram’s formula, which allows for the evalua-tion of exponential sums over non-degenerate integer-valued symmetricbilinear forms: the cited proof [7, p. 127–131] uses Fourier analysis.
Mathematics Subject Classification.
When the number field is Q , we recover the Landsberg-Schaar relation,and Milgram’s formula in this instance is essentially Lemma 1 below.Secondly, the authors of [2] suggest, in a parenthetical remark on thesecond page, that it does not seem possible to prove Hecke reciprocityby explicitly evaluating both sides. However, this does appear to workin the case of the Landsberg-Schaar relation. Indeed, in their bookon Gauss and Jacobi sums [1, Theorems 1.51, 1.52 and 1.54], Berndt,Evans and Williams give an elementary evaluation of φ ( a, b ) := a − X n =0 exp (cid:18) πin ba (cid:19) for coprime positive integral a and b . They deduce this result fromEstermann’s elementary evaluation [4] of φ ( a,
1) := a − X n =0 exp (cid:18) πin a (cid:19) for an odd positive integer a . One may then verify that1 √ a φ ( a, b ) = 12 √ b exp (cid:18) πi (cid:19) φ (4 b, − a )by evaluating both sides, and this equality is precisely the Landsberg-Schaar relation. In this argument, the hard work is contained in theevaluation of φ ( a, b ).To emphasise the fact that the Landsberg-Schaar relation is an iden-tity between Gauss sums, and to simplify the notation, we define, for a and b integers with a > a, b ) = 1 √ a a − X n =0 exp (cid:18) πin ba (cid:19) . Then the Landsberg-Schaar relation, for positive integral a and b , takesthe form Φ( a, b ) = √ i Φ(2 b, − a ) . The starting point for our proof is the following evaluation of a qua-dratic Gauss sum, given by Gauss in 1811 [5].
Lemma 1.
Let a be an integer, a ≥ . Then: Φ( a,
2) = i a = 0 mod 41 a = 1 mod 40 a = 2 mod 4 i a = 3 mod 4 . HE LANDSBERG-SCHAAR RELATION 3
A proof of Lemma 1 avoiding analytical techniques may be givenusing linear algebra [10]. Stronger results, which imply Lemma 1 (andPropositions 2 and 3 below), are also proved using elementary methodsin [1, Sections 1.3 and 1.5].One may easily check that Lemma 1 is exactly the Landsberg-Schaarrelation for b = 1. Our aim is to prove the Landsberg-Schaar relationin general by induction on the number of distinct prime factors of b .The induction step follows from the next three results, and the bulk ofthis article is spent proving the third. Lemma 2.
Let a , b and l be integers, a positive and ( a, b ) = 1 . Then: Φ( ab, l ) = Φ( a, bl )Φ( b, al ) . The proof is not difficult, but is hard to find in this form: usually l is assumed to be even, which simplifies matters considerably. Proof. As s runs from 0 to b − t runs from 0 to a − as + bt runsthrough a complete system of representatives for elements of Z /ab Z .So ( as + bt ) = g + 2 gkab + k a b for k = 0 or 1, 0 ≤ g < ab . It follows that:Φ( ab, l ) = 1 √ ab ab − X n =0 exp (cid:18) πin lab (cid:19) = ǫ Φ( a, bl )Φ( b, al ) , where ǫ = ( a or b even( − S a and b both odd,and S = { ( s, t ) | as + bt > ab } . The value of S is ( a − b − – the problem of determining S was setas a puzzle by Sylvester in [12] and solved by W. J. Curran Sharp inthe same volume. The solution runs as follows: define P ( x ) = (1 + x b + x b + · · · + x ab )(1 + x a + x a + · · · + x ba ) , and note that P ( x ) = 1 + · · · + 2 x ab + · · · + x ab , where the first dots comprise one term x g for each g of the form as + tb (we know that the coefficient of x g is 1 since a and b are coprime). BEN MOORE
Since each factor of P is a palindromic polynomial, so too is P , andit follows that the second dots comprise the same number of terms, allof coefficient 1. Therefore,(1 + a )(1 + b ) = P (1) = 4 + 2 { g < ab | g = as + tb } . Using the fact that { g < ab | g = as + tb } = ( ab − − S, the claim follows. So if a and b are both odd, then S is even, and ǫ = 1in this case too. (cid:3) The following result will not be needed until Section 4.
Lemma 3.
Suppose a , b and k are nonzero integers, a and k are pos-itive, and at least one of a or b is even. Then Φ( ka, kb ) = √ k Φ( a, b ) . Proof. Φ( ka, kb ) = 1 √ ka ka − X n =0 exp (cid:18) πin ba (cid:19) = 1 √ k k − X m =0 √ a a − X n =0 exp (cid:18) πi ( n + am ) ba (cid:19) = 1 √ k k − X m =0 exp ( iπabm )Φ( a, b )= √ k Φ( a, b ) . (cid:3) At this point, we only need one more result to prove the Landsberg-Schaar relation in Section 4.
Proposition 1.
Let p be a prime and l an integer with ( p, l ) = 1 .Then: Φ( p k , l )Φ( p k , − l ) = ( p an odd prime, k ≥ p = 2 , k ≥ . The next two sections are devoted to proving Proposition 1, whichis achieved by computing Φ( p k , l ) directly. All the results of the nexttwo sections are well-known in the literature, though apparently notall collected in one place. In particular, Proposition 2 and Proposition3 are special cases of Gauss’ evaluation of Φ( a, HE LANDSBERG-SCHAAR RELATION 5 to know the number of solutions to x = a mod p k for each a , which isthe subject of the next section. Acknowledgements.
The author is extremely grateful to Mike East-wood for his support and encouragement concerning this article, andmost especially for his firm belief that an elementary proof of theLandsberg–Schaar relation should exist! The author would also liketo thank Bruce Berndt for reading an earlier draft, Ram Murty forsome encouraging remarks, David Roberts for tracking down Gauss’original evaluation of his eponymous sums, and the anonymous refereefor suggesting valuable improvements to the article.2.
Counting solutions to x = a mod p k The first result is reminiscent of Hensel’s lemma, but is more direct.
Lemma 4.
Let p be a prime, not necessarily odd, and j > i . { x | x = kp i mod p j } = ( p i/ { x | x = k mod p j − i } i even i odd, ( k, p ) = 1 . Proof.
To dispose of the case where i is odd, note that x = kp i + lp j implies p i divides x , so p divides k . Now suppose that i is even. Define A = { x ∈ Z /p j Z | x = kp i mod p j } B = { y ∈ Z /p j − i Z | y = k mod p j − i } . The map F : A → B by x p i/ x is surjective, so to prove that | A | = p i/ | B | , we need only show that each fibre of F has cardinality p i/ . Since F ( x ) = F ( y ) if and only if x = y + tp j − i/ , the fibre of F ( x ) contains nothing more than the elements x t = x + tp j − i/ for t = 0 , . . . , p i/ −
1. But since p i/ divides x ,( x t ) = x + 2 xtp j − i/ + t p j + j − i/ mod p j = x + 2 stp i/ p j − i/ mod p j = x mod p j . So the fibre of F ( x ) is exactly the x t . (cid:3) We can now count the solutions to x = 0 mod p k . Lemma 5. { x | x = 0 mod p k } = ( p k/ k even p ( k − / k odd . BEN MOORE
Proof.
For k even, put j = k , i = k − { x | x = 0 mod p k } = p ( k − / { x | x = 0 mod p } , and { x | x = 0 mod p } = { , p, p, . . . , ( p − p } . For k odd, put j = k , i = k − { x | x = 0 mod p k } = p ( k − / { x | x = 0 mod p } = p ( k − / . (cid:3) The next two results are standard: one may consult Hecke [6, p. 47,Theorems 46a and 47] or Dickson [3, p. 13, Theorem 17].
Lemma 6.
Let p be an odd prime, j ≥ , ( k, p ) = 1 , and write (cid:0) kp (cid:1) forthe Legendre symbol. Then { x | x = k mod p j } = 1 + (cid:18) kp (cid:19) . Lemma 7.
For p = 2 and ( k,
2) = 1 : { x | x = k mod 4 } = ( k = 1 mod 40 k = 3 mod 4 . For j ≥ : { x | x = k mod 2 j } = ( k = 1 mod 80 otherwise. Lemma 4 and Lemma 6 taken together give us a complete picturefor odd p when k = 0, as follows. Lemma 8.
For ( k, p ) = 1 , j > i and p an odd prime: { x | x = kp i mod p j } = ( p i/ (cid:16) (cid:0) kp (cid:1)(cid:17) i even i odd. The analogue of Lemma 8 for p = 2 follows from Lemma 4 andLemma 7. Since the exceptional cases j = 1 and j = 2 can be done byhand, we only need to consider j − i ≥ Lemma 9.
For p = 2 , ( k,
2) = 1 , j ≥ i + 3 and i even: { x | x = 2 i k mod 2 j } = ( p i/ k = 1 mod 80 otherwise.For i odd (and all other hypotheses unchanged): { x | x = 2 i k mod 2 j } = 0 . HE LANDSBERG-SCHAAR RELATION 7 Evaluation of Φ( p k , l )This section is devoted to evaluating Φ( p k , l ) for p prime and ( l, p ) =1. We first evaluate Φ( p k , l ) in the case that p is an odd prime, thenwe proceed to the exceptional case of p = 2. The idea of each proofis to expand Φ( p k , l ) as a finite Fourier series. The coefficients havebeen calculated in Section 2, and substituting in these expressions andsimplifying yields the claimed results. In these calculations we implic-itly make use of Lemma 8, Lemma 9 and Lemma 5. We conclude thissection with the proof of Proposition 1. Proposition 2.
Let p be an odd prime and ( l, p ) = 1 . Then: Φ( p k , l ) = ( k even, k ≥ (cid:0) lp (cid:1) Φ( p k , k odd.Proof. First we treat the case of k even.Φ( p k , l ) = p − k/ p k − X n =0 { x | x = n mod p k } exp (cid:18) πinlp k (cid:19) = p − k/ p k − X n =0 (cid:18) (cid:18) np (cid:19)(cid:19) exp (cid:18) πinlp k (cid:19) + ( p k/ − − p k − − X n =1 exp (cid:18) πinplp k (cid:19) + X i =2 , ,...,k − p k − i − X n =1( n,p )= p i n/p i = m { x | x = mp i mod p k } . We should explain each term in the last two lines: the first term givesthe correct coefficients for ( n, p ) = 1, the second term makes the correctcontribution for n = 0, the third term makes the n th coefficient 0 forany nonzero n divisible by p , and the final term restores the correctcoefficient for these n . Note that the Legendre symbol (cid:0) np (cid:1) is definedto be 0 if n is divisible by p – this implies that (cid:0) · p (cid:1) is multiplicative. BEN MOORE
The inner sum in the last term can be simplified: p k − i − X n =1( n,p )= p i n/p i = m { x | x = mp i mod p k } = p i/ p k − i − X m =1 (cid:18) (cid:18) mp (cid:19)(cid:19) exp (cid:18) πimlp k − i (cid:19) − p k − i − − X m =1 (cid:18) (cid:18) mpp (cid:19)(cid:19) exp (cid:18) πimlp k − i − (cid:19) = p i/ p k − i − X m =1 (cid:18) mp (cid:19) exp (cid:18) πimlp k − i (cid:19) = p i/ (cid:18) lp (cid:19) p k − i − X m =0 (cid:18) mp (cid:19) exp (cid:18) πimp k − i (cid:19) . The final equality above follows from the facts that (cid:0) · p (cid:1) is multiplicative,and that ( l, p ) = 1 implies that as m runs from 0 to p k − i −
1, so does lm mod p k − i . But this last sum is zero, since i ≤ k − k − i > r > P p r − m =0 (cid:0) mp (cid:1) exp (cid:16) πimp r (cid:17) = 0 as follows: p r − X m =0 (cid:18) mp (cid:19) exp (cid:18) πimp r (cid:19) = p r − − X α =0 ( α +1) p X n = αp (cid:18) np (cid:19) exp (cid:18) πinp r (cid:19) = p r − − X α =0 p − X n =0 (cid:18) m + αpp (cid:19) exp (cid:18) πi ( m + αp ) p r (cid:19) = p r − − X α =0 exp (2 πiαp r − ) p − X n =0 (cid:18) mp (cid:19) exp (cid:18) πimp r (cid:19) = 0 . Therefore, the last term in the expansion of Φ( p k , l ) vanishes. Whenwe expand the factor (1 + (cid:0) np (cid:1) ) multiplying the first term, we find thatthe sum multiplied by 1 is a geometric series, so it vanishes, and thesum multiplied (cid:0) np (cid:1) is zero by the calculation above. The − p k , l ) = 1, as promised.Now we treat odd k , and suppose for the moment that k >
1. Thenthe coefficients are very similar, apart from the contributions for n = 0 HE LANDSBERG-SCHAAR RELATION 9 and n = mp k − with ( m, p ) = 0. Specifically,Φ( p k , l ) = p − k/ p k − X n =0 (cid:18) (cid:18) np (cid:19)(cid:19) exp (cid:18) πinlp k (cid:19) + ( p ( k − / − − p ( k − / − X n =1 exp (cid:18) πinplp k (cid:19) + X i =2 , ,...,k − p k − i − X n =1( n,p )= p i n/p i = m { x | x = mp i mod p k } . The calculation above shows that each inner sum in the last term van-ishes, except in the case i = k −
1, in which the condition k − i > p − X m =1 m = αp p ( k − / (1 + (cid:18) mp (cid:19) ) exp (cid:18) πimlp (cid:19) = p ( k − / − p − X m =1 (cid:18) mp (cid:19) exp (cid:18) πimlp (cid:19)! = p ( k − / (cid:18) − √ p (cid:18) lp (cid:19) Φ( p, (cid:19) . As with the case for k even, the first term in the expansion of Φ( p k , l )vanishes, the − p ( k − / (cid:18) − √ p (cid:18) lp (cid:19) Φ( p, (cid:19) , so we are left with Φ( p k , l ) = (cid:18) lp (cid:19) Φ( p, . If k = 1, then it is clear that Φ( p, l ) = (cid:0) lp (cid:1) Φ( p,
2) in this casetoo. (cid:3)
Proposition 3.
Suppose k ≥ and ( l,
2) = 1 . Then:
Φ(2 k , l ) = ( √ (cid:0) πil (cid:1) k odd (cid:0) πil (cid:1) k even Proof.
As before, for k ≥ p k , l ) = p − k/ p k − X n =0 { x | x = n mod p k } exp (cid:18) πinlp k (cid:19) = p − k/ p k − X n =1 mod 8 { x | x = n mod p k } exp (cid:18) πinlp k (cid:19) + k − X i =1 X ( n,p )= p i ,n/p i = m { x | x = mp i mod p k } exp (cid:18) πimlp i p k (cid:19) + N , where N = p k/ if k is even, and N = p ( k − / if k is odd.Suppose k ≥ p k , l ) = p − k/ p k − X n =1 mod 8 (cid:18) πinlp k (cid:19) + p k/ + p k/ exp (cid:18) πil (cid:19) + X i =2 , ,...,k − p k − i − X n =1 mod 8 p i/ exp (cid:18) πinlp i p k (cid:19) = p − k/ p k − − X α =0 (cid:18) πi (1 + p α ) lp k (cid:19) + p k/ + p k/ exp (cid:18) πil (cid:19) + X i =2 , ,...,k − p k − i − − X α =0 p i/ exp (cid:18) πil (1 + p α ) p k − i (cid:19) . Since k ≥
4, and the term i = k − p k − i − − > i = 2 , , . . . , k −
4, so the final sum is a geometric series and vanishes.Similarly, k ≥ k , l ) = 1 + exp (cid:0) πil (cid:1) .Now suppose k > i = k − HE LANDSBERG-SCHAAR RELATION 11 term corresponding to i = k − p k , l ) = p − k/ p k − − X α =0 (cid:18) πi (1 + p α ) lp k (cid:19) + p ( k − / + p ( k − / exp ( πil ) + 4 p ( k − / exp (cid:18) πil (cid:19) + X i =2 , ,...,k − p k − i − − X α =0 p i/ exp (cid:18) πil (1 + p α ) p k − i (cid:19) . Then since p k − i − − > i = 2 , , . . . , k −
5, the final sum vanishes,as does the first sum, so we are left with:Φ(2 k , l ) = p − / (cid:18) − l + 2 exp (cid:18) πil (cid:19)(cid:19) = √ (cid:18) πil (cid:19) . Lastly, suppose k = 3. Then compared to the case k > i = k − i = 0 whichis already accounted for by the first term. For ease of comparison wewrite this expression out before explicitly setting k = 3:Φ( p k , l ) = p − k/ p k − − X α =0 (cid:18) πi (1 + p α ) lp k (cid:19) + p ( k − / + p ( k − / exp ( πil ) . Now we set k = 3 in the expression above, and simplify:Φ( p k , l ) = 2 − / (cid:18) (cid:18) πil (cid:19) + 2 + 2( − l (cid:19) = √ (cid:18) πil (cid:19) . (cid:3) Finally, we can prove Proposition 1: Let p be an odd prime, k ≥ p, l ) = 1:Φ( p k , l )Φ( p k , − l ) = (cid:18) lp (cid:19)(cid:18) − lp (cid:19)(cid:0) Φ( p k , (cid:1) (by Proposition 2)= (cid:18) − p (cid:19)(cid:18) lp (cid:19) (cid:18) − p (cid:19) (by Lemma 1)= 1 . Let p = 2 and suppose k ≥
3, (2 , l ) = 1. By Proposition 3:Φ(2 k , l )Φ(2 k , − l ) = ((cid:0) (cid:0) πil (cid:1)(cid:1) (cid:0) (cid:0) − πil (cid:1)(cid:1) k even2 exp (cid:0) πil (cid:1) exp (cid:0) − πil (cid:1) k odd= 2 . Induction
By Lemma 1, the Landsberg-Schaar relation holds for b = 1. Weproceed by induction on the number of distinct prime factors of b . Weassume that Φ( a, b ) = √ i Φ(2 b, − a ) for all b with less than n prime fac-tors, and prove, using Proposition 1, that Φ( a, bp k ) = √ i Φ(2 bp k , − a )for all primes p . We may assume that ( b, p ) = 1, and also ( a, p ) = 1 byLemma 3. As usual, the case for p an odd prime is treated first.Φ( a, bp k ) = Φ( p k a, b )Φ( p k , ab ) (by Lemma 2)= √ i Φ(2 b, − p k a )Φ( p k , ab )= √ i Φ(2 bp k , − a )Φ( p k , ab )Φ( p k , − ab ) (by Lemma 2)= √ i Φ(2 bp k , − a ) . (by Proposition 1)Now for p = 2:Φ( a, b. k ) = Φ( a, k +1 b ) = Φ(2 k +1 a, b )Φ(2 k +1 , ab ) (by Lemma 2)= Φ(2 k +2 a, b )Φ(2 k +2 , ab ) (by Lemma 3)= √ i Φ(2 b, − k +2 a )Φ(2 k +2 , ab )= √ √ i Φ( b, − k +1 a )Φ(2 k +2 , ab ) (by Lemma 3)= √ √ i Φ(2 k +1 b, − a )Φ(2 k +2 , ab )Φ(2 k +1 , − ab ) (by Lemma 2)= √ i Φ(2 b. k , − a ) Φ(2 k +2 , ab )Φ(2 k +2 , − ab ) (by Lemma 3)= √ i Φ(2 b. k , − a ) . (by Proposition 1) HE LANDSBERG-SCHAAR RELATION 13
References [1] Berndt, B.C., Evans, R.J., Williams, K.S.: Gauss and Jacobi Sums. JohnWiley & Sons, Inc. (1998)[2] Boylan, H., Skoruppa, N.-P.: A quick proof of reciprocity for Hecke Gausssums. J. Number Theory. , 110–114 (2013)[3] Dickson, L. E.: Introduction to the Theory of Numbers. Dover Publications(1957)[4] Estermann, T.: On the sign of the Gaussian sum. J. London Math. Soc. ,66–67 (1945)[5] Gauss, C. F.: Summatio quarandum serierium singularium. Comment. Soc.Reg. Sci. Gottingensis (1811) May be found online at: https://gdz.sub.uni-goettingen.de/id/PPN602151724?tify={"view":"export"} [6] Hecke, E.: Lectures on the Theory of Algebraic Numbers. Springer-Verlag(1981)[7] Husemoller, D., Milnor, J.: Symmetric Bilinear Forms. Ergeb. Math. Gren-zgeb., , Springer-Verlag (1971)[8] Landsberg, G.: Zur Theorie der Gaussschen Summen und der linearen Trans-formation der Thetafunctionen. J. Reine Angew Math. , 234–253 (1893)May be found online at: https://archive.org/details/journalfrdierei97crelgoog/page/n241 [9] Murty, M.R., Pacelli, A.: Quadratic reciprocity via theta functions, vol. 1, pp.107–116. Ramanujan Math. Society Lecture Notes (2005)[10] Murty, M.R., Pathak, S.: Evaluation of the quadratic Gauss sum. MathematicsStudent , 139–150 (2017)[11] Schaar, M.: M´emoire sur la th´eorie des r´esidus quadratiques Acad. Roy. Sci.Lettres Beaux Arts Belgique , (1850) May be found online at: [12] Sylvester, J.J.: Question 7382. In: Mathematical Questions with theirsolutions, from the “Educational Times”, vol. 41, p. 21. Hodgson (1884). Maybe found online at: https://ia801409.us.archive.org/5/items/mathematicalque10millgoog/ School of Mathematical Sciences, University of Adelaide, SA 5005,Australia
E-mail address ::