All unitary perfect polynomials over F 2 with less than five distinct prime factors
aa r X i v : . [ m a t h . N T ] M a r All unitary perfect polynomials over F withless than five distinct prime factors Luis H. Gallardo - Olivier RahavandrainyDepartment of Mathematics, University of Brest,6, Avenue Le Gorgeu, C.S. 93837, 29238 Brest Cedex 3, France.e-mail : [email protected] - [email protected] 29, 2018 a) Running head: binary unitary perfect polynomials.b) Keywords: Sum of divisors, unitary divisors, polynomials, finite fields,characteristic 2 . c) Mathematics Subject Classification (2000): 11T55, 11T06.d) Corresponding author: Luis H. Gallardo.1 bstract We find all unitary perfect polynomials over the prime field F with less than five distinct prime factors. Let p be a prime number and let F q be a finite field of characteristic p andorder q. Let A ∈ F q [ x ] be a monic polynomial. We say that a divisor d of A is unitary if d is monic and gcd( d, Ad ) = 1. Let ω ( A ) denote the numberof distinct monic irreducible factors of A over F q and let σ ( A ) (resp. σ ∗ ( A ))denote the sum of all monic divisors (resp. unitary divisors) of A ( σ and σ ∗ are multiplicative functions).The analogue notion over the positive integers is the notion of unitary perfectnumbers. Only few results are known for them (see [15, 16, 19]), namely, allare even numbers, we know only five of them. Graham [16] characterizedthree of them, namely 6 , , . Goto [15] proved an explicit exponentialupper bound in k = ω ( n ) for n unitary perfect. Wall [19] improved a previousresult of Subbarao, by proving that ω ( n ) ≥ n. We call even a polynomial A with some zero in F q , and odd a polynomialthat is not even. We assume that A / ∈ F q . Since A and σ ( A ) have the same degree it follows that A divides σ ( A ) isequivalent to σ ( A ) = A. If σ ( A ) = A (resp. σ ∗ ( A ) = A ), then we say that A is a perfect (resp. unitary perfect) polynomial. We may consider the perfectpolynomials as a polynomial analogue of the multiperfect numbers. E. F.Canaday, the first doctoral student of Leonard Carlitz, began in 1941 [5] thestudy of perfect polynomials by working on the prime field F . Later, in theseventies, J. T. B. Beard Jr. et al. extended this work in several directions(see e.g. [2], [1], [4]) including the study of unitary perfect polynomials.We became interested in this subject a few years ago and obtain someresults ([6], [7], [8], [9], [10], [11], [12] and [13]) including for q ∈ { , } acomplete classification of the perfect polynomials A for which ω ( A ) is small.We began the study of unitary perfect polynomials by considering thesplitting case when q = p (see [14]). In this paper we study more generalunitary perfect polynomials A improving on previous results of Beard et al.[3] and Beard [2]. In particular we prove that A must be even, contrary toperfect polynomials for which we do not know whether or not there exist odd2erfect polynomials. More precisely, we determine here all unitary perfectpolynomials A , over F , such that ω ( A ) ≤
4. As usual N denotes the non-negative integers and N ∗ the positive integers.Our main results are the following:Let A be a nonconstant polynomial over F such that ω ( A ) ≤
4, then A is unitary perfect if and only if either A or A ( x + 1) is of the form B n forsome n ∈ N where: − if ω ( A ) ≤ B = x ( x + 1) ,B = x ( x + 1) ( x + x + 1) ,B ( x ) ∈ { x ( x + 1) ( x + x + 1) , x ( x + 1) ( x + · · · + x + 1) }− if ω ( A ) = 4: i ) B = x ( x + 1) (1 + x + x ) (1 + x + x ) ,ii ) B = x ( x + 1) (1 + x + x ) (1 + x + · · · + x ) ,iii ) B = x ( x + 1) (1 + x + · · · + x ) (1 + x + · · · + x ) ,iv ) B = x ( x + 1) (1 + x + x ) (1 + x + x ) ,v ) B = x ( x + 1) (1 + x + · · · + x ) (1 + x + x + x + x ) ,vi ) B = x ( x + 1) (1 + x + x )(1 + x + x ) ,vii ) B = x ( x + 1) (1 + x + x ) (1 + x + x ) ,viii ) B = x ( x + 1) (1 + x + x ) (1 + x + · · · + x ) ,ix ) B = x ( x + 1) (1 + x + x )(1 + x + · · · + x ) ,x ) B = x ( x + 1) (1 + x + x ) (1 + x + · · · + x ) ,xi ) B = x ( x + 1) (1 + x + x ) (1 + x + x ) ,xii ) B = x ( x + 1) (1 + x + x ) (1 + x + x ) . We may consider the family { x n ( x + 1) n : n ∈ N } as an analogue of thefamily { x n +1 ( x + 1) n +1 } of trivial even perfect polynomials over F . Note that Beard [2] and Beard et al. [3] computed the above list with theexception of v), x), and xi) that are new.Moreover, compared to the list of all perfect polynomials A over F with ω ( A ) < S ( x ) = 1 + x + x , S ( x + 1) ,S ( x ) = 1 + x + x + x + x , S ( x + 1) S ( x ) = 1 + x + · · · + x , S ( x + 1) ,S ( x ) = 1 + x + · · · + x , S ( x + 1) . It is clear from the above results that the classification of all perfector unitary perfect polynomials A with a moderately large number ω ( A ) ofdistinct prime factors may become very complicated. New tools need to bediscovered to make more progress in this area. We need the following results. Some of them are obvious, so we omit to givetheir proofs. Our first result give information on the sizes of the primaryparts of unitary perfect polynomials.
Lemma 2.1. (see also [2, Theorem 1]) If A = P h · · · P h r r Q k · · · Q k s s is anonconstant unitary perfect polynomial over F q such that: (cid:26) P , . . . , P r , Q , . . . , Q s are both irreducible h deg( P ) = · · · = h r deg( P r ) < k deg( Q ) ≤ · · · ≤ k s deg( Q s ) . Then: r ≡ p ) . Proof.
By definition, one has: 0 = σ ∗ ( A ) − A = AP h + · · · + AP h r r + · · · In particular, r = 1+ · · · +1, which is the leading coefficient of AP h + · · · + AP h r r ,equals 0 in F p . Lemma 2.2. If A = A A is unitary perfect over F and if gcd( A , A ) = 1 .Then A is unitary perfect if and only if A is unitary perfect. Lemma 2.3. If A ( x ) is unitary perfect over F , then the polynomials A ( x +1) and A n are also unitary perfect over F , for any n ∈ N . We recall here some useful notation and results in Canaday’s paper [5]:4
We define as the inverse of a polynomial P ( x ) of degree m , the poly-nomial P ∗ ( x ) = x m P ( 1 x ). • We say that P inverts into itself if P = P ∗ . • A polynomial P is complete if P = 1 + x + · · · + x h , for some h ∈ N .Part iii) of the following lemma is essentially a result of Dickson (see [5,Lemma 2]) Lemma 2.4 (see [5, lemma 7], [11, Lemma 2.1]) . i) Any complete polynomialinverts into itself. ii) If x + · · · + x h = P Q , where
P, Q are irreducible,then either ( P = P ∗ , Q = Q ∗ ) or ( P = Q ∗ , Q = P ∗ ) . iii) If P = P ∗ , P irreducible and if P = x a ( x + 1) b + 1 , then: P ∈ { x + x , x + · · · + x } . Lemma 2.5. (see [5, Lemmata 4, 5, 6 and Theorem 8])
Let
P, Q ∈ F [ x ] such that P is irreducible and let n, m ∈ N . i) If P + · · · + P n = Q m , then m ∈ { , } . ii) If P + · · · + P n = Q m A , with m > and A ∈ F [ x ] is nonconstant,then deg( P ) > deg( Q ) . iii) If x + · · · + x n = P Q and P = 1 + ( x + 1) + · · · + ( x + 1) m , then n = 4 , P = 1 + x + x and Q = P ( x ) = 1 + x + x . iv) If any irreducible factor of x + · · · + x n is of the form x a ( x + 1) b + 1 ,then n ∈ { , , } . v) If x + · · · + x h = 1 + ( x + 1) + · · · + ( x + 1) h , then h = 2 n − , forsome n ∈ N . Lemma 2.6. If x + x divides x + · · · + x h , then h ≡ .If x + · · · + x divides x + · · · + x h , then h ≡ . As a special case of [17, Theorem 2.47], we have
Lemma 2.7.
The polynomial x + · · · + x m is irreducible over F if andonly if: m + 1 is a prime number and is a primitive root in F m +1 . Consequently one gets 5 emma 2.8. i) The polynomial Q ( x ) = 1 + x + · · · + ( x ) l is irreducibleover F if and only if l = 4 . ii) The polynomial Q ( x ) = 1 + x + · · · + x . r is irreducible over F if andonly if r = 2 . iii) The polynomial Q ( x ) = 1 + x + · · · + x . r is irreducible over F if andonly if r = 1 .Proof. We prove only necessity. Sufficiency is obtained by direct computa-tions.i): For k ∈ N ∗ , let Φ k be the k -th cyclotomic polynomial over F . Recall thatif k is a prime number, then Φ k ( x ) = 1 + x + · · · + x k − .If Q ( x ) is irreducible, then 1 + x + · · · + x l is also irreducible.Thus, by Lemma 2.7, l + 1 is a prime number and Q ( x ) = Φ l +1 ( x ).It remains to observe that if 5 = l + 1, then:Φ l +1 ( x ) = Φ l +1 ( x ) Φ l +1) ( x ) . So that Q is not irreducible in that case. We conclude that l = 4.ii): If Q ( x ) is irreducible, then by Lemma 2.7, p = 3 . r + 1 is a primenumber and 2 is a primitive root in F p . So, 2 is not a square in F p . Byconsidering the Legendre Symbol ( 2 p ) = ( − p − , we see that we must have r ∈ { , } .The case r = 1 does not happen since Q ( x ) is irreducible.iii): As above, we obtain: r ∈ { , } . The case r = 2 does not happen since5 . r + 1 is prime.We prove now the non-existence of odd unitary perfect polynomials: Lemma 2.9.
Any nonconstant unitary perfect polynomial over F is divisibleby x and by x + 1 . In particular, there is no odd unitary perfect polynomialover F .Proof. If P is an odd prime polynomial over F , then P (0) = P (1) = 1,so that for any positive integer h , 1 + P (0) h = 1 + P (1) h = 0. Thus, themonomials x and x + 1 divide 1 + P h . Now, let A be an unitary perfectpolynomial. We have ω ( A ) ≥
2. If both x, x + 1 divide A , then we are done.If there exists an odd polynomial P ∈ F [ x ] such that P h | A and P h +1 ∤ A ,then σ ∗ ( P h ) = 1 + P h divides σ ∗ ( A ) = A . So x, x + 1 divide A .6 emark 2.10. • In the rest of the paper, we put S ( x ) = S ( x + 1) for S ∈ F [ x ] . • For Theorems 3.1 and 4.1, we shall prove only necessity, since suffi-ciency is always obtained by direct computations. ω ( A ) ≤ We prove the following result:
Theorem 3.1.
Let A ∈ F [ x ] be a polynomial such that ω ( A ) ≤ , then A is unitary perfect over F if and only if either A or A is of the form B n for some n ∈ N , where: i ) B = x + x,ii ) B ∈ { x ( x + 1) ( x + x + 1) , x ( x + 1) ( x + · · · + x + 1) } ,iii ) B = x ( x + 1) ( x + x + 1) . ω ( A ) = 2 The following proposition gives the first part of Theorem 3.1.
Proposition 3.2.
Let A ∈ F [ x ] such that ω ( A ) = 2 , then A is unitaryperfect over F if and only if A is of the form ( x + x ) n , for some n ∈ N .Proof. It remains to prove necessity since sufficiency is obvious.The case where A ∈ { x h P k , ( x +1) h P k } , with P odd, is impossible by Lemma2.9. So A splits: A = x h ( x + 1) k . We must have: 1 + x h = ( x + 1) h , x +1) k = x k . Hence, h = k = 2 n , for some n ∈ N .Consequently the unitary perfect polynomials A with ω ( A ) = 2 are ex-actly the perfect polynomials with ω ( A ) = 2 . ω ( A ) = 3 In this case, A is of the form x h ( x + 1) k P l , with P odd. Lemma 3.3. If A = x h ( x + 1) k P l is an unitary perfect polynomial over F ,then l = 2 n , for some nonnegative integer n . roof. Put: l = 2 n u , where u is odd and n ∈ N . Since the only prime divisorsof A = σ ∗ ( A ) are x, x + 1 and P , and since P does not divide 1 + P l , thepolynomial 1 + P l = σ ∗ ( P l ) must be of the form x a ( x + 1) b . Thus,(1 + P )(1 + P + · · · + P u − ) = 1 + P u = x c ( x + 1) d . Since x, x + 1 divide 1 + P and since gcd(1 + P, P + · · · + P u − ) = 1, weconclude that u − h = 2 h c, k = 2 k d with c, d odd. Since A is unitary perfect, we have x h = ( x + 1) h (1 + x + · · · + x c − ) h , x + 1) k = x k (1 + ( x + 1) + · · · + ( x + 1) d − ) k , P n = (1 + P ) n = ( x a ( x + 1) b ) n . (1)Lemma 2.5-i) implies that:1 + x + · · · + x c − , x + 1) + · · · + ( x + 1) d − ∈ { , P } . Since h and k play symmetric roles and since P must appear in the righthand side of (1), we may reduce the study to the two cases:(I) : 1 + x + · · · + x c − = P, d = 1 , (II) : 1 + x + · · · + x c − = P = 1 + ( x + 1) + · · · + ( x + 1) d − . According to Lemma 2.4-iii), we have: P ∈ { x + x , x + · · · + x } and c ∈ { , } .By considering exponents and degrees, System (1) implies k = h + 1 , n = h if c = 3 ,k = h + 2 , n = h if c = 5 . We obtain part ii) of Theorem 3.1.
We have c = d and P = P . So, by Lemma 2.4, P = 1 + x + x , and hence c = d = 3. System (1) implies: k = h, n = h + 1, and we obtain part iii) ofTheorem 3.1. This completes the proof of Theorem 3.1.It turns out that we can also get Theorem 3.1. as a consequence of a niceresult of Swan: 8 .2.3 Another proof using Swan’s Lemma We would like to give, here, another proof of parts ii) and iii) of Theorem 3.1,by using Lemma 2.1 and the following result about reducibility of a binarypolynomial in F [ x ]: Lemma 3.4 (see [18], p. 1103, line 3) . Let n, k ∈ N be such that n > k ,then the polynomial x n + x k + 1 is reducible over F . From that, we obviously obtain the
Corollary 3.5.
Let r be a positive integer, then the polynomial P = x r + x r − + 1 is irreducible over F if and only if r ∈ { , } . We recall that A is of the form x h ( x + 1) k P l , with P odd and l = 2 n forsome n ∈ N . Put p = deg( P ). By Lemma 2.1, we have either ( h = k ≤ lp )or ( h = lp ≤ k ) or ( k = lp ≤ h ). The third case is similar to the secondsince h and k play symmetric roles.Case h = k ≤ lp We obtain A = x h ( x + 1) h P n , h ≤ n p . Since A is unitary perfect,we have 1 + x h = ( x + 1) b P c , x + 1) h = x a P c , P n = (1 + P ) n = ( x a ( x + 1) b ) n . Hence: P = x a ( x + 1) b + 1 , ( x + 1) b P c = 1 + x h = 1 + ( x + 1 + 1) h = ( x + 1) a ( P ( x + 1)) c . It follows that: a = b , c = c ≥ , P ( x ) = P ( x + 1) . Thus, c = c = 2 n − and a = b . The irreducibility of P implies a = b = 1.So, P = x + x + 1. Put h = 2 h c , where c is odd. We have now:(1 + x ) h (1 + x + · · · + x c − ) h = 1 + x h = ( x + 1) b ( x + x + 1) n − . c = 3 and h = n −
1. We get A = B n − , where B = x ( x + 1) ( x + x + 1) . So we obtain part iii) of Theorem 3.1.Case h = lp ≤ k We obtain now: A = x h ( x + 1) k P n , h = 2 n p ≤ k . Since A is unitaryperfect, we have 1 + x h = (1 + x p ) n = (( x + 1) b P c ) n , x + 1) k = x a P c , P n = (1 + P ) n = ( x a ( x + 1) b ) n . Hence: a + c p = k , b + c p = p, n c + c = 2 n . It follows that c ∈ { , } . If c = 0, then b = p and 1 + x p = ( x + 1) p , so p = 2 r , for some r ∈ N ∗ . Thus, a + b = 2 r . Since P = x a ( x + 1) b + 1 isirreducible, a and b must be both odd. Moreover, c = 2 n and a + 2 n r = a + c p = k = 2 n ( b + b ) = 2 n (2 r + b ) . Hence a = 2 n b , and (1 + ( x + 1) r + b ) n = 1 + ( x + 1) k = x a P c = ( x b P ) n . It follows that:1 + ( x + 1) r + b = x b P = x b ( x a ( x + 1) b + 1) . Thus, b = 1 , a = 2 r − , k = 2 n (2 r + 1) , P = x r − ( x + 1) + 1 , and A = ( x r ( x + 1) r +1 P ) n . So by Corollary 3.5, we get r ∈ { , } and A satisfies part ii) of Theorem 3.1.If c = 1, then c = b = 0. It follows that 1 + x p = P , with p ≥
2. Thiscontradicts the fact that P is irreducible.10 Case ω ( A ) = 4 We prove the following result:
Theorem 4.1.
Let A ∈ F [ x ] be a polynomial such that ω ( A ) = 4 , then A is unitary perfect over F if and only if either A or A is of the form B n for some n ∈ N , where: i ) B = x ( x + 1) (1 + x + x ) (1 + x + x ) ,ii ) B = x ( x + 1) (1 + x + x ) (1 + x + · · · + x ) ,iii ) B = x ( x + 1) (1 + x + · · · + x ) (1 + x + · · · + x ) ,iv ) B = x ( x + 1) (1 + x + x ) (1 + x + x ) ,v ) B = x ( x + 1) (1 + x + · · · + x ) (1 + x + x + x + x ) ,vi ) B = x ( x + 1) (1 + x + x )(1 + x + x ) ,vii ) B = x ( x + 1) (1 + x + x ) (1 + x + x ) ,viii ) B = x ( x + 1) (1 + x + x ) (1 + x + · · · + x ) ,ix ) B = x ( x + 1) (1 + x + x )(1 + x + · · · + x ) ,x ) B = x ( x + 1) (1 + x + x ) (1 + x + · · · + x ) ,xi ) B = x ( x + 1) (1 + x + x ) (1 + x + x ) ,xii ) B = x ( x + 1) (1 + x + x ) (1 + x + x ) . The following proposition gives more details about the form of an unitaryperfect polynomial.
Proposition 4.2.
Every unitary perfect polynomial A over F , with ω ( A ) =4 , is of the form x h ( x + 1) k P l u Q m , where:i) P, Q, u are odd, deg( P ) ≤ deg( Q ) , ii) h , k ∈ N ∗ , l, m ∈ N and either ( u = 1) or ( u = 3 , Q = 1 + P + P ) ,iii) P ∈ { x + x , x + · · · + x } if P is complete , iv) deg( Q ) ≥ if Q is complete . Proof.
First of all, x and x + 1 divide A by Lemma 2.9. So A = x h ( x + 1) k P r Q s , for some h , k , r, s ∈ N ∗ . Put r = 2 l u, s = 2 m v , where u, v are odd and l, m ∈ N . Consider σ ∗ ( Q s ) = 1 + Q s = (1 + Q ) m (1 + Q + · · · + Q v − ) m . x and x + 1 divide 1 + Q , they do not divide 1 + Q + · · · + Q v − .Hence, 1 + Q + · · · + Q v − ∈ { , P } , by Lemma 2.5-i). If v − ≥
2, then1 + Q + · · · + Q v − = P . This is impossible because deg( P ) ≤ deg( Q ).Thus, v − s = 2 m . Now, by considering degrees, we see thatthe irreducible odd polynomial Q does not divide 1 + P . It follows that(1 + P ) l (1 + P + · · · + P u − ) l = 1 + P r = σ ∗ ( P r ) must be of the form x a ( x + 1) b Q c . Thus, by Lemma 2.5-i):1 + P + · · · + P u − ∈ { , Q } . We conclude that either ( u = 1) or (1 + P + · · · + P u − = Q ).If u >
1, then put u = 2 w + 1. We get1 + Q m = (1 + Q ) m = ( P (1 + P + · · · + P u − ) ) m = ( P (1 + P ) ( P + · · · + P w − ) ) m . Since x, x + 1 and P divide 1 + Q and since x, x + 1 divide 1 + P , noneof the irreducible divisors of A does divide 1 + P + · · · + P w − . Hence w = 1, u = 3 and Q = 1 + P + P . Since deg( P ) ≤ deg( Q ), the irreduciblepolynomial Q does not divide 1+ P . So P is always of the form x a ( x +1) b +1.If P is complete, then by parts i) and iii) of Lemma 2.4, we have P ∈{ x + x , x + · · · + x } . Finally, if Q is complete, since 1+ x + x is the onlydegree 2 odd irreducible polynomial over F , we must have deg( Q ) ≥ p = deg( P ) , q = deg( Q ) , h = 2 h c, k = 2 k d, with c, d odd . Since A is unitary perfect and since Q does not divide 1 + P , we have: x h = (1 + x c ) h = (1 + x ) h (1 + x + · · · + x c − ) h = (1 + x ) h P h c Q h d , x + 1) k = x k (1 + (1 + x ) + · · · + (1 + x ) d − ) k = x k P k c Q k d , P l u = (1 + P ) l (1 + P + · · · + P u − ) l = ( x a (1 + x ) b ) l Q l d , Q m = (1 + Q ) m = ( x a (1 + x ) b P c ) m . (2)By considering degrees and exponents of x, x + 1 , P and Q , (2) implies: h c = 2 h (1 + pc + qd ) = 2 k + 2 l a + 2 m a , k d = 2 k (1 + pc + qd ) = 2 h + 2 l b + 2 m b , l up = 2 l ( a + b + qd ) = (2 h c + 2 k c + 2 m c ) p, m q = 2 m ( a + b + pc ) = (2 h d + 2 k d + 2 l d ) q. (3)12y Lemma 2.5, c , d , c , d , d ∈ { , } so that:1 + x + · · · + x c − , x ) + · · · + (1 + x ) d − ∈ { , P, Q, P Q } . Since h and k play symmetric roles, and since x, x + 1 , P and Q must divide A = σ ∗ ( A ), it is sufficient to consider the following ten cases:(I) : c = d = 1 , (II) : 1 + x + · · · + x c − = P, d = 1 , (III) : 1 + x + · · · + x c − = Q, d = 1 , (IV) : 1 + x + · · · + x c − = P Q, d = 1 , (V) : 1 + x + · · · + x c − = P = 1 + ( x + 1) + · · · + ( x + 1) d − , (VI) : 1 + x + · · · + x c − = Q, x + 1) + · · · + ( x + 1) d − = P, (VII) : 1 + x + · · · + x c − = P Q, x + 1) + · · · + ( x + 1) d − = P, (VIII) : 1 + x + · · · + x c − = Q = 1 + ( x + 1) + · · · + ( x + 1) d − , (IX) : 1 + x + · · · + x c − = P Q, x + 1) + · · · + ( x + 1) d − = Q, (X) : 1 + x + · · · + x c − = P Q = 1 + ( x + 1) + · · · + ( x + 1) d − . In this case, if u = 1, then since Q must appear in the right hand sideof System (2), Q must divide 1 + P , which is impossible. So, u = 3 and1+ Q = P ( P +1). Thus, System (2) implies that c = 1 and 3 · l = c · m = 2 m so that 3 divides 2 m . It is impossible. As above, u = 3 and Q = 1 + P + P . By Proposition 4.2, we get P ∈ { x + x , x + · · · + x } and c ∈ { , } . If P = 1 + x + · · · + x , then: Q = 1 + P + P = 1 + x + x + x + x = (1 + x + x )(1 + x + x + x + x ) , which is reducible.So we must have: P = 1 + x + x . Thus, c = 3 and Q = 1 + x + x . System(3) implies that: l = m, h = m + 1 , k = m + 2 . We obtain part i) of Theorem 4.1. 13 .3 Case (III) P must divide 1 + Q since it must appear in the right hand side of (2).Put: c − r s , with s odd. We get x a (1 + x ) b +1 P c = (1 + x )(1 + Q ) = x ( x + 1)(1 + x + · · · + x c − ) . Thus, a = 1 and( x +1) b +1 P c = (1+ x )(1+ x + · · · + x c − ) = 1+ x c − = (1+ x ) r (1+ x + · · · + x s − ) r . We conclude that: b = 2 r − , c = 2 r , P = 1 + x + · · · + x s − . By Proposition 4.2, we get P ∈ { x + x , x + · · · + x } . Thus, c ∈ { · r + 1 , · r + 1 } , and by Lemma 2.8, c ∈ { , } . It followsthat we must have u = 1 , d = 0 ,P = 1 + x + x , Q = 1 + x + · · · + x if c = 13 ,P = 1 + x + · · · + x , Q = 1 + x + · · · + x if c = 11 . System (3) implies m = h, l = h + 2 , k = h + 3 if c = 13 ,m = h, l = h + 1 , k = h + 3 if c = 11 . We obtain parts ii) and iii) of Theorem 4.1.
We get 1 + x + · · · + x c − = P Q, and by Lemma 2.4: P ∈ { P ∗ , Q ∗ } . P = P ∗ In this case, by Lemma 2.4-iii), we have: P ∈ { x + x , x + · · · + x } . • If P = 1 + x + x , then by Lemma 2.5-iii), the only possibility is c = 9 , Q = 1 + x + x . u = 1 . System (3) implies the following: m = h, l = h + 1 , k = h + 2 . We obtain then part iv) of Theorem 4.1. • If P = 1 + x + · · · + x , then 1 + x + · · · + x divides 1 + x + · · · + x c − .So, by Lemma 2.6, c is divisible by 5. Put c = 5 w . We get Q = 1 + x + x + · · · + ( x ) w − = 1 + P + P . Thus, by Lemma 2.8-i) and by Proposition4.2, we have c = 5 w = 25 , u = 1 , P = 1 + x + · · · + x , Q = 1 + x + x + x + x . System (3) implies m = h, l = h + 2 , k = h + 4 . So we obtain part v) of Theorem 4.1. P = Q ∗ We get p = q . So both P and Q are of the form x a ( x + 1) b + 1. We concludeby Lemma 2.5-iv) that: c = 7 , P, Q ∈ { x + x , x + x } . It follows that Q = 1 + P + P and u = 1. System (3) implies l = m = h, k = h + 2 . We obtain then part vi) of Theorem 4.1.
In this case, by Lemma 2.4-iii), P = 1 + x + x and c = d = 3. Moreover, u must be equal to 3. So, Q = 1 + P + P = 1 + x + x . System (3) impliesnow: l = m = k = h. Consequently we obtain part vii) of Theorem 4.1.15 .6 Case (VI)
In this case, P ∈ { x + x , x + · · · + x } by Lemma 2.5-iv). So Q = 1 + P + P and hence u = 1. P does not divide Q In this case, both P and Q are of the form x a ( x + 1) b + 1. By Lemma 2.5-iv)and Proposition 4.2-iii)-iv), we have two possibilities: P = P = 1 + x + x , Q = 1 + x + · · · + x ,P = 1 + x + · · · + x = Q. Thus ( c, d ) ∈ { (5 , , (5 , } . System (3) implies m = h, l = k = h + 1 if c = 5 , d = 3 ,l = m = k = h if c = d = 5 . We obtain parts viii) and ix) of Theorem 4.1. P divides Q In this case, P must divide 1 + Qx = 1 + x + · · · + x c − . Moreover, accordingto System (2), we have a = 1 , x + · · · + x c − = ( x + 1) b P c . Thus, if we put c − r s , with s odd, we obtain(1 + x ) r (1 + x + · · · + x s − ) r = (1 + x s ) r = 1 + x c − = ( x + 1) b +1 P c . We conclude that: b = 2 r − , and by Lemma 2.5-i): P = 1 + x + · · · + x s − , c = 2 r . Hence, by Lemmata 2.5-v) and 2.4-iii)the only possiblity that remains is P = 1 + x + x = P, s = 3 , c = 3 · r + 1 . It follows that r = 2 by Lemma 2.8. System (3) implies that: m = h, k = h + 2 , l = h + 3 . We obtain part x) of Theorem 4.1. 16 .7 Case (VII)
In this case, P divides 1 + x + · · · + x c − . By Lemma 2.5-iii), we get c = 9 , P = 1 + x + x , d = 3 , Q = 1 + x + x . Moreover, u = 1 since Q = 1 + P + P .System (3) implies that: m = h, k = h + 1 , l = h + 2 . We obtain part xi) of Theorem 4.1.
In this case, by Lemma 2.5-v) and by Proposition 4.2-iv), c = d = 2 w − ≥ P must appear in the right hand side of (2) , it must divide 1 + Q = x (1 + x + · · · + x c − ). Hence P divides 1 + x + · · · + x c − . Thus, a = 1 and ( x + 1) b P c = 1 + x + · · · + x c − = (1 + x )(1 + x + · · · + x w − − ) . We deduce that: b = 1 , c = 2 , P = 1 + x + · · · + x w − − . By Proposition 4.2-iii), we must have2 w − − ∈ { , } . So w = 3 and Q = 1 + x + · · · + x = (1 + x ) which is not irreducible. In this case, Q divides 1 + x + · · · + x c − . By Lemma 2.5-iii), we get Q = 1 + x + x , P = 1 + x + x . This contradicts the fact: deg( P ) ≤ deg( Q ). In this case, by Lemma 2.5-v), by Proposition 4.2-iv) and by Lemma 2.4-ii),we get c = d = 2 w − ≥ , and either ( P = P ∗ , Q = Q ∗ ) or ( P = Q ∗ ) . .10.1 Case where P = P ∗ , Q = Q ∗ We have by Lemma 2.4-iii): P ∈ { x + x , x + · · · + x } . • If P = 1 + x + x , by Lemma 2.5-iii), Q = 1 + x + x . Thus, c = 9 = 2 w − • If P = 1 + x + · · · + x , then P divides 1 + x + · · · + x d − . So, by Lemma2.5, d − P = Q ∗ We have p = q and both P, Q are of the form x a ( x + 1) b + 1. By Lemma2.5-iv), c = d = 7 and P, Q ∈ { x + x , x + x } . Moreover u = 1, by Proposition 4.2-ii). System (3) implies that: l = m = h + 1 , k = h. We obtain finally part xii) of Theorem 4.1. This completes the proof of theTheorem.
We are grateful to the referee of a first version of this paper for carefulreading and for suggestions that improved the presentation of the paper.We are including in the next section his report (but excluding the detailedtechnical suggestions to authors).
Referee report on the paper ”All unitary perfect polynomials over F2 withless than five distinct prime factors” by Luis H. Gallardo and Olivier Raha-vandrainy.The authors are studying the problem of finding all the unitary perfectpoly nomials over finite fields. The present paper contains the full classifica-tion of all the perfect unitary polynomials over F2 and serves as a continu-ation of a series of their publication devoted to the same topic. Previously18he problem was studied by E.F. Canaday, J.T.B. Beard Jr, A.T. Bulloc,M.S. Harbin, J.R. Oconnel Jr, K.I. West. The latest publication on studyof the perfect unitary polynomials was published in 1991, and this makespapers of the mentioned authors hardly available. Moreover, publications[2]-[4] in the reference list is unavailable since the journal Rend. Acad. Lin-cei they published in has status ”no longer indexed” in database of the AMSand the journal’s webpage containing the mentioned volumes was not found.Happily the authors are citing the papers [2]-[4] only in the history of thequestion. The general idea of the proofs of the results in the paper is ratherele- mentary. But it requires a great scope of computations and applies moredeep results on irreducibility of the polynomials. Some of these irreducibil-ity results was proved by the authors in their previous papers. In generalthe paper makes good impression by numerous tricks used by the authors tosimplify computations. The paper worth to be published in the Journal, itpresents a new research which devoted to an interesting problem. The au-thors gives several interesting ideas, combination of which solves a problem.I found several misprints and places where arguments of proofs are not clear.I’d like to recommend the authors to correct misprints and clarify uncleararguments in proofs.
From the (seemingly favorable ?) report above it was deduced that thepreliminary version of this paper was not suitable for publication in theIJNT.
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