Factorization of Cyclotomic Polynomial Values at Mersenne Primes

Home , Cyclotomic polynomial

arXiv:2106.10008v1 [math.NT] 18 Jun 2021 a trzto fccooi oyoilvalues polynomial cyclotomic of Factorization )MteaisSbetCasfiain(00:1T5 11T06. 11T55, (2010): Classification Subject Mathematics b) )Kyod:Ccooi oyoil,smo iios nt fields finite divisors, of sum polynomials, Cyclotomic Keywords: a) )Crepnigato:O Rahavandrainy O. author: Corresponding c) -al:ligl@nvbetf [email protected] - [email protected] : e-mail aoaor eMt´mtqe eBean Atlantique Bretagne Mathématiques de de Laboratoire alroLi .-RhvnriyOlivier Rahavandrainy - H. Luis Gallardo nvri´ eBet M NS6205 CNRS UMR Brest, Université de tMren primes Mersenne at ue2,2021 21, June 1 r Abstract We get some results about the factorization of φp(M) ∈ F2[x], where p is a prime number, φp is the corresponding cyclotomic polynomial and M is a Mersenne prime (polynomial). By the way, we better understand the factorization of the sum of the divisors of M 2h, for a positive integer h.

1 Introduction

We define a Mersenne prime (polynomial) over F2 as an irreducible polynomial of the form 1 + xa(x + 1)b, for some positive integers a, b. In that case: gcd(a, b)=1and (a or b is odd). For a nonzero polynomial A ∈ F2[x], let ω(A) (resp. σ(A)) denote the number of all its distinct irreducible factors (resp. the sum of all its divisors). Recall that σ is a multiplicative function. If σ(A)= A, then we say that A is perfect. In order to characterize all perfect polynomials over F2, Conjecture 1.2 plays an important role. That is why we try to prove it, since a few years (see e.g., [1], [2] and [3]). Let φp be the cyclotomic polynomial associated to a prime number p. We p−1 p−1 see that: φp(M)=1+ M + ··· + M = σ(M ). Therefore, (new) results about the factorization of φp(M) would be useful to prove Conjecture 1.2. It is convenient to fix some notations. Notations • For S ∈ F2[x], we denote by: - S the polynomial obtained from S with x replaced by x+1: S(x)= S(x+1), s−l - αl(S) the coefficient of x in S, 0 ≤ l ≤ s. One has: α0(S) = 1. - N (resp. N∗) the set of nonnegative integers (resp. of positive integers). • For a fixed prime Mersenne M := 1 + xa(x + 1)b and for a prime number p := 2h + 1, we set:

2h 2h 2h−1 U2h := σ(σ(M )), W := WM = U2h + σ(M )+1, R := RM = σ(M )+ W, degM := deg(M)= a + b, c := cM =2h degM − deg(W ).

By Corollary 2.7, we consider, for p ≥ 5, the following sets of Mersenne

2 primes:

1 2 Σp := {M : cM = degM }, Σp := {M : cM = degM +1,cM odd}, 3 Σp := {M : cM + mM = degM ,cM even}, 4 Σp := {M : cM + mM = degM +1,cM even and mM ≥ 3}, 1 2 3 4 and Σp := Σp ∪ Σp ∪ Σp ∪ Σp. We shall prove the following result: Theorem 1.1. Let p ≥ 5 be a prime number. Then, for any M 6∈ Σp, p−1 φp(M)= σ(M ) is divisible by a non Mersenne prime. Theorem 1.1 is a new step on the proof of a Conjecture about Mersenne primes that is discussed in recent papers ([2], [3]): Conjecture 1.2 (Conjecture 5.2 in [1]). Let h ∈ N∗ and M be a Mersenne prime such that (M 6∈ {1+ x + x3, 1+ x2 + x3} or h ≥ 2). Then, the polynomial σ(M 2h) is divisible by a non Mersenne prime.

j Remark 1.3. Conditions deﬁning the sets Σp are very restrictive (see Table below to illustrate this fact). For several “small” values of p and of d, the j j set Λp,d := {M ∈ Σp :5 ≤ deg(M) ≤ d} is empty or of small cardinality. We denote by Md the set of Mersenne primes M such that 5 ≤ deg(M) ≤ d.

1 2 3 4 p d #Λp,d #Λp,d #Λp,d #Λp,d #Md 5 100 4 0 0 0 226 7 100 0 0 2 2 226 11 100 2 0 2 0 226 13 100 0 2 0 0 226 17 100 2 0 0 0 226 19 100 0 0 0 0 226 23 100 0 0 0 0 226 29 100 4 0 2 0 226 53 60 2 0 4 0 138 59 60 0 0 0 0 138 61 60 2 2 0 0 138 67 60 0 0 0 0 138 71 60 0 2 0 0 138

We hope that it remains a few things to completely establish that Conjecture.

3 We mainly prove Theorem 1.1, by contradiction (to Corollary 2.4 or to Lemma 2.5-iii)). Precisely, we shall prove that there exists an odd positive integer ℓ such that αℓ(U2h)=1or αℓ(R) = 1.

2 Useful facts

In Section 3, we shall use Lemmas 2.1, 2.2, 2.8, 2.11 and Corollary 2.12 (sometimes, without explicit mentions).

Lemma 2.1. Let S ∈ F2[x] of degree s ≥ 1 and l,t,r,r1,...,rk ∈ N such that r1 > ··· >rk, t ≤ k,r1 − rt ≤ l ≤ r ≤ s. Then r1 rk i) αl[(x + ··· + x )S]= αl(S)+ αl−(r1−r2)(S)+ ··· + αl−(r1−rt)(S). ii) αl(σ(S)) = αl(S) if no irreducible polynomial of degree at most r divides S.

Proof. i) (resp. ii)) follows from the deﬁnition of αl (resp. from the fact: σ(S)= S + T , where deg(T ) ≤ deg(S) − r − 1).

2h 2h Lemma 2.2. One has: αl(σ(M )) = αl(M ) if 1 ≤ l ≤ a + b − 1, 2h 2h 2h−1 αl(σ(M )) = αl(M + M ) if a + b ≤ l ≤ 2(a + b) − 1.

Proof. Since σ(M 2h)= M 2h + M 2h−1 + T , with deg(T ) ≤ (a + b)(2h − 2) = 2h 2h 2h(a + b) − 2(a + b), Lemma 2.1-ii) implies that αl(σ(M )) = αl(M ) if 1 ≤ 2h 2h 2h−1 l ≤ a+b−1 and αl(σ(M )) = αl(M +M ) if a+b ≤ l ≤ 2(a+b)−1.

Lemma 2.3. [2, Lemmas 4.6 and 4.8] i) σ(M 2h) is square-free and it is not a Mersenne prime. ii) a ≥ 2 or b ≥ 2 so that M =1+6 x + x2.

We may assume, by Lemma 2.3, that:

2h aj bj σ(M )= Y Mj, Mj =1+ x (x + 1) irreducible, Mi =6 Mj if i =6 j. (1) j∈J

Corollary 2.4. i) The integers u = X aj and v = X bj are both even. j∈J j∈J ii) U2h splits (over F2). iii) U2h is a square so that αk(U2h)=0 for any odd positive integer k.

4 Proof. i): See [2, Corollary 4.9]. ii) and iii): Assumption (1) implies:

2h aj bj u v U2h = σ(σ(M )) = σ(Y Mj)= Y x (x + 1) = x (x + 1) , j∈J j∈J with u and v both even.

2h Lemma 2.5. i) degM ≥ 5 and ω(σ(M )) ≥ 3. ii) W =06 and it is not a square. iii) R = σ(M 2h−1)+ W is a square and M 2h−1 + W is not a square. iv) If deg(W ) is even, then there exists a (least) odd integer m ≥ 1 such that αm(W )=1. v) If deg(W ) is odd, then there exists a (least) even integer e ≥ 1 such that αe(W )=1.

2h Proof. i): The case: degM ≤ 4 or ω(σ(M )) ≤ 2, is already treated in [3]. 2h 2h ii): If 0 = W = U2h + σ(M ) + 1, then σ(M ) = U2h + 1 is a square by Corollary 2.4-iii). It contradicts i). 2h If W is a square, then σ(M ) = U2h +1+ W is also a square, which is impossible. 2h−1 2h iii): R = σ(M )+ W = U2h +1+ M which is a square. If M 2h−1 + W = R + σ(M 2h−2) is a square, then σ(M 2h−2) is a square, with 2h − 2 ≥ 2. It is impossible. iv): follows from the fact that W is not a square. deg(W ) v): If for any even e, αe(W ) = 0, then W + x is a square. So, by diﬀerentiating: 0 = (W + xdeg(W ))′ = W ′ + xdeg(W )−1. We get the contradiction: xdeg(W )−1 = W ′ =(R + σ(M 2h−1))′ =0+(σ(M 2h−1))′ =(σ(M 2h−1))′.

Corollary 2.6. If c < a + b (resp. c > a + b), then c (resp. a + b) is even.

Proof. - If c < a + b, then deg(W ) > deg(σ(M 2h−1)) and 2h(a + b) − c = deg(W ) = deg(R) is even. - If c > a + b, then deg(W ) < deg(σ(M 2h−1)) and (2h − 1)(a + b) = deg(σ(M 2h−1)) = deg(R) is even.

Corollary 2.7. If c = a + b +1 (resp. c + m = a + b, c + m = a + b +1 with m ≥ 3), then c is odd (resp. even, even).

5 Proof. - If c = a + b + 1, then c > a + b. So, a + b is even. - If c + m = a + b, then c < a + b. So, c is even. - If c + m = a + b + 1 with m ≥ 3, then c < a + b. So, c is even.

2h Lemma 2.8. i) For any odd integer l ≥ 1, αl(M )=0. 2h−1 ii) If b is odd, then α1(M )=1. Proof. i): because M 2h is a square. 2h−1 (2h−1)b (2h−1)b (2h−1)b ii): α1(M ) = α1(x · (x + 1) ) = α1((x + 1) ) = 1, since (2h − 1)b is odd. By direct computations, we get:

Lemma 2.9. Let S, T ∈ F2[x] such that deg(S) = deg(T ). If deg(S) is even (resp. odd), then S + T is a square if and only if for any odd (resp. even) ε, αε(S)= αε(T ).

Lemma 2.10. Let S, T ∈ F2[x] such that deg(S) = deg(T ). Then deg(S + T ) = deg(S) − ℓ where ℓ is the least integer such that αℓ(S) =6 αℓ(T ).

Lemma 2.11. Let S, T ∈ F2[x], with s = deg(S) > t = deg(T ). Then: - For any l

2h 2h−1 Corollary 2.12. i) If a + b is odd, then αa+b(M + M )=1. 2h 2h−1 ii) If a + b is even, then αa+b+1(M + M )=1. 2h−1 iii) If a + b is even, then α1(σ(M ))=1. Proof. We apply Lemma 2.11 with S = M 2h, T = M 2h−1. We get s = 2h(a + b) and t = (2h − 1)(a + b) so that s − t = a + b. Remark that for any odd integer l, αl(S)=0, since S is a square. i) if a + b is odd, then αa+b(S + T )= αa+b(S)+ α0(T )=0+1=1. 2h 2h−1 ii) if a + b is even, then a and b are both odd. Thus, αa+b+1(M + M )= 2h−1 αa+b+1(S)+ α1(T )=0+ α1(M ) = 1, since b is odd (Lemma 2.8-ii)). 2h−1 2h−1 iii): b is odd and α1(σ(M )) = α1(M ) = 1 (Lemma 2.8-ii)).

~~2h Corollary 2.13. The integer c is the least one such that αc(U2h)+αc(σ(M )) = 1.~~

~~2h Proof. Apply Lemma 2.10 with S = U2h and T = σ(M ) + 1, since deg(S + T ) = deg(W )=2h deg(M) − c = deg(S) − c.~~

~~6 3 The proof~~

~~We should consider four cases, since M 6∈ Σp. But, Corollary 2.7 implies that Case IV does not happen.~~

~~Case I: c ≥ a + b +2 Case II: c + m < a + b Case III: c < a + b~~

~~3.1 I: c ≥ a + b +2 Lemma 3.1. One has: i) αa+b(U2h)=1 if a + b is odd. ii) αa+b+1(U2h)=1 if a + b is even.~~

~~2h Proof. Write: U2h =(σ(M ))+(W +1) = S +T , with deg(S)−deg(T )= c. i): One has: a + b, a + b +1~~

3.2 II: c + m < a + b Since c < a + b, Corollary 2.6 implies that c and thus deg(W ) are even. We consider the odd integer m (of Lemma 2.5) such that: αm(W ) = 1. We recall that : 2h 2h u v W =1+ σ(M )+ U2h =1+ σ(M )+ x (x + 1) , R = σ(M 2h−1)+ W is a square. One has: c + m =2h degM − deg(W )+ m is odd.

~~Lemma 3.2. αc+m(U2h)=1.~~

~~2h 2h−1 Proof. Write: U2h = S + T with S = M and T = W + σ(M ) + 1. One has: deg(S) − deg(T )=2h(a + b) − deg(W )= c. Lemma 2.11 implies:~~

~~2h αc+m(U2h)= αc+m(M )+ αm(T )=0+ αm(T )= αm(T ).~~

~~7 But, m < a + b − c = deg(W ) − deg(σ(M 2h−1) + 1). Again, from Lemma 2.11, one has: αm(T )= αm(W ) = 1. So, αc+m(U2h) = 1.~~

~~3.3 III: c < a + b < c + m As above, c and thus deg(W ) are even.~~

Lemma 3.3. If c + m > a + b and a + b odd, then αa+b−c(R)=1. Proof. Set: R = S + T with S = W and T = σ(M 2h−1). One has: deg(S) − deg(T )= a + b − c and a + b − c < m. Therefore, 2h−1 αa+b−c(W ) = 0 and αa+b−c(R)= αa+b−c(W )+α0(σ(M )) = 0+1 = 1.

Lemma 3.4. If c + m > a + b +1 and a + b even, then αa+b−c+1(R)=1. Proof. As above, set: S = W and T = σ(M 2h−1). One has: deg(S) − deg(T )= a + b − c and a + b − c +1 < m. 2h−1 By Corollary 2.12, α1(σ(M )) = 1. 2h−1 So, αa+b−c+1(R)= αa+b−c+1(W )+ α1(σ(M )) = 0 + 1 = 1.

~~References~~

[1] L. H. Gallardo, O. Rahavandrainy, On even (unitary) perfect polynomials over F2 , Finite Fields Appl. 18 (2012), 920–932. [2] L. H. Gallardo, O. Rahavandrainy, On Mersenne polynomials over F2, Finite Fields Appl. 59 (2019), 284–296. [3] L. H. Gallardo, O. Rahavandrainy, On (unitary) perfect polynomials over F2 with only Mersenne primes as odd divisors, arXiv: 1908.00106v1 (Jul. 2019).