On (unitary) perfect polynomials over $\mathbb{F}_2$ with only Mersenne primes as odd divisors

TL;DR

This paper characterizes all (unitary) perfect polynomials over the finite field _2 with only Mersenne primes as odd divisors, revealing that only nine such polynomials exist, based on new factorization results.

Contribution

It provides a complete classification of certain perfect polynomials over _2 involving Mersenne primes, introducing new factorization results for polynomials of the form M^{2h+1} +1.

Findings

Only nine (unitary) perfect polynomials over _2 are formed from x, x+1, and Mersenne primes.

New factorization results for M^{2h+1} +1 over _2.

Implications for the structure of odd perfect polynomials over _2.

Abstract

The only (unitary) perfect polynomials over $\mathbb{F}_2$ that are products of $x$, $x+1$ and Mersenne primes are precisely the nine (resp. nine "classes") known ones. This follows from a new result about the factorization of $M^{2h+1} +1$, for a Mersenne prime $M$ and for a positive integer $h$. Other consequences of such a factorization are new results about odd perfect polynomials.

Equations

$$\begin{array}[]{l}T_{1}=x^{2}(x+1)M_{1},T_{2}=\overline{T_{1}},\\ T_{3}=x^{4}(x+1)^{3}M_{3},T_{4}=\overline{T_{3}},\ T_{5}=x^{4}(x+1)^{4}M_{3}\overline{M_{3}}=\overline{T_{5}},\\ T_{6}=x^{6}(x+1)^{3}M_{2}\overline{M_{2}},T_{7}=\overline{T_{6}},\\ T_{8}=x^{4}(x+1)^{6}M_{2}\overline{M_{2}}M_{3}\text{ and }T_{9}=\overline{T_{8}},\\ U_{1}=x^{3}(x+1)^{3}M_{1}^{2},\ U_{2}=x^{3}(x+1)^{2}M_{1},\ U_{3}=x^{5}(x+1)^{4}M_{3},\\ U_{4}=x^{7}(x+1)^{4}M_{2}\overline{M_{2}},\ U_{5}=x^{5}(x+1)^{6}M_{1}^{2}M_{3},\ U_{6}=x^{5}(x+1)^{5}M_{3}\overline{M_{3}},\\ U_{7}=x^{7}(x+1)^{7}M_{2}^{2}{\overline{M_{2}}}^{2},\ U_{8}=x^{7}(x+1)^{6}M_{1}^{2}M_{2}\overline{M_{2}},\ U_{9}=x^{7}(x+1)^{5}M_{2}\overline{M_{2}}\ \overline{M_{3}}.\\ \end{array}$$

$$\sigma (M^{2h})=\prod _{j\in J}{P}_j$$

$$\#{\mathcal{M}}_m\le \#\{c:1\le c\le m,\gcd (c,m)=1\}=\phi (m).$$

$$U_{2h}=\sigma (\sigma (M^{2h}))=\sigma (\prod _{j\in J}{P}_j)=\prod _{j\in J}{x}^{a_j}(x+1{)}^{b_j}=x^u(x+1{)}^v,$$

$$\begin{array}[]{l}0=\alpha_{1}(M^{2h})=\alpha_{1}(\sigma(M^{2h}))=\alpha_{1}(B)+1,\\ \alpha_{3}(U_{2h})=\alpha_{3}(\sigma(B))+\alpha_{2}(\sigma(B))=\alpha_{3}(B)+\alpha_{2}(B),\\ 0=\alpha_{3}(M^{2h})=\alpha_{3}(\sigma(M^{2h}))=\alpha_{3}(B)+\alpha_{2}(B)+\alpha_{1}(B).\end{array}$$

$$\sigma (M^2)=M_1\cdots M_r,r\ge 3\text{ and }W:=U_4=\sigma (\sigma (M^2)).$$

$$\begin{array}[]{l}i)\ \alpha_{1}(\sigma(M^{2}))=\alpha_{1}(B)+1,\ \alpha_{2}(\sigma(M^{2}))=\alpha_{2}(B)+\alpha_{1}(B)+1,\\ ii)\ \alpha_{3}(\sigma(M^{2}))=\alpha_{3}(B)+\alpha_{2}(B)+\alpha_{1}(B),\\ iii)\ \alpha_{3}(\sigma(M^{2}))=0.\par\end{array}$$

$${\alpha }_1(W)={\alpha }_1(B)+1,{\alpha }_2(W)={\alpha }_2(B)+{\alpha }_1(B),{\alpha }_3(W)={\alpha }_3(B)+{\alpha }_2(B).$$

$$\begin{array}[]{l}\alpha_{1}(W)=\alpha_{1}(\sigma(B))+1=\alpha_{1}(B)+1,\\ \alpha_{2}(W)=\alpha_{2}(\sigma(B))+\alpha_{1}(\sigma(B))=\alpha_{2}(B)+\alpha_{1}(B),\end{array}$$

Full text

On (unitary) perfect polynomials over $\mathbb{F}_{2}$ with only Mersenne primes as odd divisors

Gallardo Luis H. - Rahavandrainy Olivier

Université de Brest, UMR CNRS 6205

Laboratoire de Mathématiques de Bretagne Atlantique

e-mail : luisgall@univ-brest.fr - rahavand@univ-brest.fr

• a)

Running head: Mersenne and perfect polynomials

• b)

Keywords: Sum of divisors, polynomials, finite fields, characteristic $2.$

• c)

Mathematics Subject Classification (2010): 11T55, 11T06.

• d)

Corresponding author:

Luis H. Gallardo

**Abstract

**The only (unitary) perfect polynomials over $\mathbb{F}_{2}$ that are products of $x$, $x+1$ and Mersenne primes are precisely the nine (resp. nine “classes”) known ones. This follows from a new result about the factorization of $M^{2h+1}+1$, for a Mersenne prime $M$ and for a positive integer $h$. Other consequences of such a factorization are new results about odd perfect polynomials.

1 Introduction

Let $A\in\mathbb{F}_{2}[x]$ be a nonzero polynomial. We say that $A$ is even if it has a linear factor and that it is odd otherwise. We define a Mersenne prime (polynomial) over $\mathbb{F}_{2}$ as an irreducible polynomial of the form $1+x^{a}(x+1)^{b}$, for some positive integers $a,b$. This comes as an analogue of the prime factors of the even perfect numbers. As over the integers, we say that a divisor $d$ of $A$ is unitary if $\displaystyle{\gcd(d,\frac{A}{d})=1}$. Let $\omega(A)$ denote the number of distinct irreducible (or prime) factors of $A$ over $\mathbb{F}_{2}$ and let $\sigma(A)$ (resp. $\sigma^{*}(A)$) denote the sum of all (unitary) divisors of $A$ (both $\sigma$ and $\sigma^{*}$ are multiplicative functions). If $\sigma(A)=A$ (resp. $\sigma^{*}(A)=A$), then we say that $A$ is (unitary) perfect. Finally, we say that a (unitary) perfect polynomial is indecomposable if it is not a product of two coprime nonconstant (unitary) perfect polynomials.

The notion of (unitary) perfect polynomials is introduced in [6] (a simplified version of this Ph. D. thesis under Carlitz) by E. F. Canaday in $1941$ and extended by J. T. B. Beard Jr. et al. (probably still inspired by Carlitz that advised the advisor of Beard) in several directions ([2], [4]). Later research in the subject ([8], [10], [12], [14], [16]) allows us to more precisely describe such polynomials. For the perfect case, we get:

• the “trivial” ones, of the form $(x^{2}+x)^{2^{n}-1}$, for some positive integer $n$,

• nine others which are the unique even all whose odd factors are Mersenne primes raised to powers of the form $2^{n}-1$ ([18], Theorem 1.1),

• and the last two which are divisible by a non Mersenne prime.

By analogy, since we can also consider perfect polynomials, $A\in\mathbb{F}_{2}[x]$ with $\sigma(A)/A=1$, as an analogue of multiperfect numbers, $n\in\mathbb{N}^{*}$ avec $\sigma(n)/n\in\mathbb{N}^{*}$, it might have some interest to observe that most known multiperfect numbers (see OEIS sequence A007691) appear to be divisible by a Fermat prime or by a Mersenne prime.

Obviously, all unitary perfect polynomials are even. We prove for the unitary case that essentially, the known ones belong to the nine “classes” relative to the equivalence relation : two unitary perfect polynomials are equivalent if and only if some power of $2$ of one equals some power of $2$ of the other (see below).

The paper consists of two major results that we describe now. The most important is Theorem 1.2 that improves significantly on these results (because, now there are no conditions on the powers of the $M_{j}$’s). Its proof is obtained from new results given in Theorem 1.4 which in turn, extends recent non-trivial results in [22, Theorem 1.4 ].

We began to study odd perfect polynomials in [8]. They are all squares [6] and must have [10] at least five distinct prime divisors. We have also considered [8] “special perfect” polynomials which are of the form $S=P_{1}^{2}\cdots P_{m}^{2}$, with each $P_{j}$ odd and irreducible. We proved [8] that if such a polynomial $S$ is perfect, then $\omega(S)\geq 10$, $\min_{j}{\deg(P_{j})}\geq 30$ and $P_{j}\equiv 1\mod x^{2}+x+1$. We get a new result for them as well as a new result about the existence of more general odd perfect polynomials, in Theorem 1.3, as a consequence of Theorem 1.4.

Observe that Theorem 1.4 is a new step on the proof of a Conjecture about Mersenne primes that is discussed in the recent paper [22].

It is convenient to fix some notations:

• (a)

For $S\in\mathbb{F}_{2}[x]$, we denote by $\overline{S}$ the polynomial obtained from $S$ with $x$ replaced by $x+1$: $\overline{S}(x)=S(x+1)$.

• (b)

$\mathbb{N}$ (resp. $\mathbb{N}^{*}$) denotes, as usual, the set of nonnegative integers (resp. of positive integers).

• (c)

To avoid trivialities, we suppose that any (unitary) perfect polynomial is indecomposable.

Notations 1.1**.**

*Set $M_{j}:=1+x(x+1)^{j},j\in\{1,2,3\}$, ${\mathcal{M}}:=\{M_{1},M_{2},\overline{M_{2}},M_{3},\overline{M_{3}}\}$,

${\mathcal{P}}:=\{T_{1},\ldots,T_{9}\}$ and ${\mathcal{P}}_{u}:=\{U_{1},\ldots,U_{9}\}$ where:*

[TABLE]

*The nine nontrivial perfect polynomials cited above are: $T_{1},\ldots,T_{9}$ and the two others are: $T_{10}=x^{2}(x+1)(x^{4}+x+1){M_{1}}^{2}$, $T_{11}=\overline{T_{10}}$.

The known unitary perfects are all of the form $B^{2^{n}}$, where $n\in\mathbb{N}$ and $B\in\{U_{1},\ldots,U_{9}\}$.*

Our results are:

Theorem 1.2**.**

Let $A=x^{a}(x+1)^{b}\prod_{i}P_{i}^{h_{i}}\in\mathbb{F}_{2}[x]$ with each $P_{i}$ Mersenne prime and $h_{i}\in\mathbb{N}^{*}$. Then $A$ is even $(unitary)$ perfect if and only if $A\in{\mathcal{P}}$ $(resp.\ A=B^{2^{n}}$ with $n\in\mathbb{N}$ and $B\in{\mathcal{P}}_{u})$.

Theorem 1.3**.**

*i) There exists no special perfect polynomial divisible only by Mersenne primes.

ii) If $A=P_{1}^{2h_{1}}\cdots P_{m}^{2h_{m}}$, where each $P_{j}$ is a Mersenne prime and if for some $j$, $2h_{j}+1$ is divisible by a Mersenne prime $\not=7$ or by a Fermat prime $\not=5$, then $A$ is not perfect.*

Theorem 1.4**.**

*Let $h$ be a positive integer and $M\in\mathbb{F}_{2}[x]$ a Mersenne prime. Then, in the following cases, $\sigma(M^{2h})$ is divisible by a non Mersenne prime:

i) $(M\in\{M_{1},M_{3},\overline{M_{3}}\})$ or $(M\in\{M_{2},\overline{M_{2}}\}$ and $h\geq 2)$.

ii) $M\not\in{\mathcal{M}}$ and $2h+1$ is divisible by a prime number $p$, where $(p\not=7$ is a Mersenne number$)$ or $($the order of $2$ modulo $p$ is divisible by $8$ $($in particular, when $p$ is a Fermat prime greater than $5))$.*

2 Proofs of Theorems 1.2 and 1.3

Sufficiency in Theorem 1.2 is obtained by direct computations. For the necessity, we shall apply Lemmas 2.3 and 2.9, Propositions 2.7 and 2.10. We use Theorem 1.4 to prove these two propositions. A similar method gives Theorem 1.3. We recall below [22, Theorem 1.4 ] which partially solves [18, Conjecture 5.2] about the factorization of $\sigma(M^{2h})$:

Conjecture 2.1** (Conjecture 5.2 in [18]).**

Let $h\in\mathbb{N}^{*}$ and $M$ be a Mersenne prime over $\mathbb{F}_{2}$ such that $M\not\in\{M_{2},\overline{M_{2}}\}$. Then, the polynomial $\sigma(M^{2h})$ is divisible by a non Mersenne prime.

Lemma 2.2** (Theorem 1.4 in [22]).**

Let $h\in\mathbb{N}^{*}$ such that $p=2h+1$ is prime, $M$ a Mersenne prime such that $M\not\in\{M_{2},\overline{M_{2}}\}$ and $\omega(\sigma(M^{2h}))=2$. Then, $\sigma(M^{2h})$ is divisible by a non Mersenne prime.

2.1 Proof of Theorem 1.2

We set $\displaystyle{A:=x^{a}(x+1)^{b}\prod_{i\in I}P_{i}^{h_{i}}=A_{1}A_{2}\in\mathbb{F}_{2}[x]}$, where $a,b,h_{i}\in\mathbb{N}$, $P_{i}$ is a Mersenne prime, $\displaystyle{A_{1}=x^{a}(x+1)^{b}\prod_{P_{i}\in{\mathcal{M}}}P_{i}^{h_{i}}}$ and $\displaystyle{A_{2}=\prod_{P_{j}\not\in{\mathcal{M}}}P_{j}^{h_{j}}}$.

We suppose that $A$ is indecomposable (unitary) perfect.

2.1.1 Case of perfect polynomials

Lemma 2.3** (Theorem 1.1 in [18]).**

If $h_{i}=2^{n_{i}}-1$ for any $i\in I$, then $A\in{\mathcal{P}}$.

We get from Theorem 8 in [6] and from Theorem 1.4:

Lemma 2.4**.**

*i) If $\sigma(x^{a})$ is divisible only by Mersenne primes, then $a\in\{2,4,6\}$ and all its divisors lie in ${\mathcal{M}}$.

ii) Let $M\in{\mathcal{M}}$ such that $\sigma(M^{a})$ is divisible only by Mersenne primes, then $a=2$ and $M\in\{M_{2},\overline{M_{2}}\}$.*

Lemma 2.5**.**

If $P$ is a Mersenne prime divisor of $\sigma(A_{1})$, then $P,\overline{P}\in\{M_{1},M_{2},M_{3}\}$.

Proof.

We apply Lemma 2.4. If $P$ divides $\sigma(x^{a})\cdot\sigma((x+1)^{b})$, then $P\in{\mathcal{M}}$. If $P$ divides $\sigma(P_{i}^{h_{i}})$ with $P_{i}\in{\mathcal{M}}$, then $P_{i}\in\{M_{2},\overline{M_{2}}\}$ and $P,\overline{P}\in\{M_{1},M_{3}\}$. ∎

Lemma 2.6**.**

i) For any $P_{j}\not\in{\mathcal{M}}$, one has: $\gcd(P_{j}^{h_{j}},\sigma(A_{1}))=1$ and $h_{j}=0$. ii) $A=A_{1}$.

Proof.

i): Let $P_{j}\not\in{\mathcal{M}}$ and $Q_{i}\in{\mathcal{M}}$. Then, $P_{j}$ divides neither $\sigma(x^{a})$, $\sigma((x+1)^{b})$ nor $\sigma(Q_{i}^{h_{i}})$. Thus $\gcd(P_{j}^{h_{j}},\sigma(A_{1}))=1$.

Observe that $P_{j}^{h_{j}}$ divides $\sigma(A_{2})$ because $P_{j}^{h_{j}}$ divides $A=\sigma(A)=\sigma(A_{1})\sigma(A_{2})$. Hence, $A_{2}$ divides $\sigma(A_{2})$. So, $A_{2}$ is perfect and it is equal to $1$ from the indecomposibility of $A$.

ii) follows from i). ∎

Proposition 2.7**.**

If $A_{1}$ is perfect, then $h_{j}=2^{n_{j}}-1$ for any $P_{j}\in{\mathcal{M}}$.

Proof.

i) Suppose that $P_{j}\not\in\{M_{2},\overline{M_{2}}\}$. If $h_{j}$ is even, then $\sigma(P_{j}^{h_{j}})$ is divisible by a non Mersenne prime $Q$. So, we get the contradiction: $Q\mid A$. If $hj=2^{n_{j}}u_{j}-1$ with $u_{j}\geq 3$ odd, then $\sigma(P_{j}^{h_{j}})=(1+P_{j})^{2^{n_{j}}-1}\cdot(1+P_{j}+\cdots+P_{j}^{u_{j}-1})^{2^{n_{j}}}$ is also divisible by a non Mersenne prime, which is impossible.

ii) If $P_{j}\in\{M_{2},\overline{M_{2}}\}$ and ($h_{j}$ is even or it is of the form $2^{n_{j}}u_{j}-1$, with $u_{j}\geq 3$ odd), then $a,b\in\{7\cdot 2^{n}-1:n\geq 0\}$. Thus, for some $\nu\in\mathbb{N}^{*}$, $M_{1}^{2^{\nu}}$ divides $\sigma(A)=A$. It is impossible by the part i) of our proof. ∎

Lemma 2.6, Proposition 2.7 and Lemma 2.3 imply

Corollary 2.8**.**

One has: $A=A_{1}\in{\mathcal{P}}$.

2.1.2 Case of unitary perfect polynomials

Similar proofs give Proposition 2.10 and thus, our result.

Lemma 2.9** (Theorem 1.3 in [18]).**

If $h_{i}=2^{n_{i}}$ for any $i\in I$, then $A$ (or $\overline{A}$) is of the form $B^{2^{n}}$ where $B\in{\mathcal{P}}_{u}$.

Proposition 2.10**.**

*i) If $A_{1}$ is unitary perfect then $h_{j}=2^{n_{j}}$ for any $P_{j}\in{\mathcal{M}}$.

ii) $A=A_{1}$.*

Remark 2.11**.**

Contrary to our proofs in the present paper, the proofs of [18, Corollaries 5.3 and 5.4 ] are not complete, since the special case where $\gcd(M_{2}\overline{M_{2}},A)\neq 1$ was not considered*.*

2.2 Proof of Theorem 1.3

We also use in this proof Lemma 2.2 and Theorem 1.4.

If $S=P_{1}^{2}\cdots P_{m}^{2}$ is perfect, where each $P_{j}$ is Mersenne, then $\sigma(P_{1}^{2})\cdots\sigma(P_{m}^{2})=\sigma(S)=S=P_{1}^{2}\cdots P_{m}^{2}$. We must have: $\sigma(P_{1}^{2})=\prod_{k}Q_{k}$.Thus, $P_{1}\in\{M_{2},\overline{M_{2}}\}$ and any $Q_{k}\in\{M_{1},M_{3},\overline{M_{3}}\}$. But, for any $T\in\{M_{1},M_{3},\overline{M_{3}}\}$, $\sigma(T^{2})$ and thus $S$ is divisible by a non Mersenne prime, which is impossible.

We get in the same manner the part ii) of the theorem.

3 Proof of Theorem 1.4

We mainly prove Theorem 1.4 by contradiction (to Corollary 3.5). Lemma 3.2 states that $\sigma(M^{2h})$ is square-free for any $h\in\mathbb{N}^{*}$. We suppose that:

[TABLE]

We set $U_{2h}:=\sigma(\sigma(M^{2h}))$ and $M:=x^{a}(x+1)^{b}+1$, with $M$ irreducible (so that $\gcd(a,b)=1$, $a$ or $b$ is odd). We may assume that $a$ is odd, without loss of generality.

3.1 Useful facts

Some of the following results are obvious or cited in [22], so we omit their proofs. By Lemma 3.7, $\sigma(M^{2h})$ is divisible by a non Mersenne prime whenever $\sigma(M^{p-1})$ is too, for some prime divisor $p$ of $2h+1$.

Lemma 3.1**.**

*For $m\in\mathbb{N}^{*}$, denote, as usual, by $N_{2}(m)$ the number of irreducible polynomials of degree $m$, over $\mathbb{F}_{2}$. Then

i) $N_{2}(m)\geq[2^{m}-2(2^{m/2}-1)]/m$.

ii) $\varphi(m)<N_{2}(m)$ if $m\geq 4$, where $\varphi$ is the Euler totient function.

iii) There exists at least one irreducible polynomial of degree $m$ which is not a Mersenne prime, if $m\geq 4$.*

Proof.

i): See Exercise 3.27, p. 142 in [24].

ii): We get by direct computations, $m<(2^{m}-2(2^{m/2}-1))/m$ for $4\leq m\leq 5$ and by studying the function $f(x)=2^{x}-2(2^{x/2}-1)-x^{2}$, for $x\geq 6$. So, $\varphi(m)\leq m<N_{2}(m)$.

iii): First, $1+x^{c}(x+1)^{d}\text{ Mersenne prime}$ implies that $\gcd(c,c+d)=\gcd(c,d)=1$. Moreover, the set ${\mathcal{M}}_{m}$ of Mersenne primes of degree $m$ is a subset of $\Sigma_{m}:=\{x^{c}(x+1)^{m-c}+1:1\leq c\leq m,\ \gcd(c,m)=1\},$ Thus,

[TABLE]

Therefore, there exist at most $\varphi(m)$ Mersenne primes of degree $m$. So, we get iii). ∎

Lemma 3.2**.**

*i) $\sigma(M^{2h})$ is square-free and reducible.

ii) $a\geq 2$ or $b\geq 2$ so that $M\not=M_{1}$.*

Notation 3.3**.**

For a nonconstant polynomial $S$ of degree $s$, we denote by $\alpha_{l}(S)$ the coefficient of $x^{s-l}$ in $S$, $0\leq l\leq s.$ One has: $\alpha_{0}(S)=1$.**

We sometimes apply Lemmas 3.4 and 3.6 without explicit mentions.

Lemma 3.4**.**

*Let $S\in\mathbb{F}_{2}[x]$ of degree $s\geq 1$ and $l,t,r,r_{1},\ldots,r_{k}\in\mathbb{N}$ such that $r_{1}>\cdots>r_{k}$, $t\leq k,r_{1}-r_{t}\leq l\leq r\leq s.$ Then

i) $\alpha_{l}[(x^{r_{1}}+\cdots+x^{r_{k}})S]=\alpha_{l}(S)+\alpha_{l-(r_{1}-r_{2})}(S)+\cdots+\alpha_{l-(r_{1}-r_{t})}(S)$.

ii) $\alpha_{l}(\sigma(S))=\alpha_{l}(S)$ if no irreducible polynomial of degree at most $r$ divides $S$.*

Proof.

i): Obvious, by definition of $\alpha_{l}$.

ii) Follows from the fact: $\sigma(S)=S+T$, where $\deg(T)\leq\deg(S)-r-1$. ∎

Corollary 3.5**.**

*i) The integers $\displaystyle{u=\sum_{j\in J}a_{j}}$ and $\displaystyle{v=\sum_{j\in J}b_{j}}$ are both even.

ii) $U_{2h}$ splits $($over $\mathbb{F}_{2})$.

iii) $U_{2h}$ is a square so that $\alpha_{k}(U_{2h})=0$ for any odd positive integer $k$.*

Proof.

i): See [22, Corollary 4.9].

ii) and iii): Assumption (1) implies that

[TABLE]

with $u$ and $v$ both even. ∎

Lemma 3.6**.**

One has modulo $2$: $\alpha_{l}(\sigma(M^{2h}))=\alpha_{l}(M^{2h})$ if $1\leq l\leq a+b-1$, $\alpha_{l}(\sigma(M^{2h}))=\alpha_{l}(M^{2h}+M^{2h-1})$ if $a+b\leq l\leq 2(a+b)-1$.

Proof.

Since $\sigma(M^{2h})=M^{2h}+M^{2h-1}+T$, with $\deg(T)\leq(a+b)(2h-2)=2h(a+b)-2(a+b)$, Lemma 3.4-ii) implies that $\alpha_{l}(\sigma(M^{2h}))=\alpha_{l}(M^{2h})$ if $1\leq l\leq a+b-1$ and $\alpha_{l}(\sigma(M^{2h}))=\alpha_{l}(M^{2h}+M^{2h-1})$ if $a+b\leq l\leq 2(a+b)-1$. ∎

Lemma below (with analogous proof) is a generalization of Lemma 4.10 in [22].

Lemma 3.7**.**

If $k$ divides $2h+1$ $($with $k$ prime or not$)$, then $\sigma(M^{k-1})$ divides $\sigma(M^{2h})$.

We fix a prime factor $p$ of $2h+1$. We denote by $ord_{p}(2)$ the order of $2$ in $\mathbb{F}_{p}\setminus\{0\}$.

Lemma 3.8**.**

For any $j\in J$, $ord_{p}(2)$ divides $a_{j}+b_{j}=\deg(P_{j})$.

Proof.

Let $d=\gcd_{i}(a_{i}+b_{i})$. By Lemma 4.13 in [22], $p$ divides $2^{d}-1$. Thus, $ord_{p}(2)$ divides $d$. ∎

Lemma 3.9**.**

*Let $P_{i}=1+x^{a_{i}}(x+1)^{b_{i}}$ be a prime divisor of $\sigma(M^{p-1}),$ where $2^{a_{i}+b_{i}}-1=p_{i}$ is a prime number. Then

i) any irreducible polynomial $($Mersenne or not$)$ of degree $a_{i}+b_{i}$ divides $\sigma(M^{p-1})$.

ii) $\sigma(M^{p-1})$ is divisible by a non Mersenne prime if $a_{i}+b_{i}\geq 4$.*

Proof.

First, $P_{i}$ is a primitive polynomial. Let $\alpha$ be a root of $P_{i}$. One has $M(\alpha)^{p}=1$, $M(\alpha)=\alpha^{r}$ for some $1\leq r\leq p_{i}-1$. Thus, $1=M(\alpha)^{p}=\alpha^{rp}$, with $ord(\alpha)=p_{i}$. So, $p_{i}$ divides $rp$ and $p_{i}=p$.

i): If $P$ is an irreducible polynomial of degree $a_{i}+b_{i}$, then $P$ is primitive. Let $\beta$ be a root of $P$. One has $ord(\beta)=p_{i}=p$, $P(\beta)=0$ and $M(\beta)=\beta^{s}$, for some $1\leq s\leq p_{i}-1$. Thus, $M(\beta)^{p}=\beta^{ps}=1$.

ii) follows from i) and from Lemma 3.1-iii). ∎

Corollary 3.10**.**

For any $i\in J$, $a_{i}+b_{i}\leq 3$ or $2^{a_{i}+b_{i}}-1$ is not prime.

Lemma 3.11**.**

Let $P,Q\in\mathbb{F}_{2}[x]$ such that $\deg(P)=r$, $2^{r}-1$ is prime, $P\nmid Q(Q+1)$ but $P\mid Q^{p}+1$. Then $2^{r}-1=p$.

Proof.

Let $\beta$ be a root of $P$. $\beta$ is primitive, $ord(\beta)=2^{r}-1$, $Q(\beta)\not\in\{0,1\}$ because $P\nmid Q(Q+1)$. Thus, $Q(\beta)=\beta^{t}$ for some $1\leq t\leq 2^{r}-2$. Hence, $1=Q(\beta)^{p}=\beta^{tp}.$ So, $2^{r}-1$ divides $tp$ and $2^{r}-1=p$. ∎

Corollary 3.12**.**

Let $r\in\mathbb{N}^{*}$ such that $2^{r}-1$ is a prime distinct from $p$. Then, no irreducible polynomial of degree $r$ divides $\sigma(M^{p-1})$.

Proof.

If $P$ divides $\sigma(M^{p-1})$ with $\deg(P)=r$, then $P$ divides $M^{p}+1$ and by taking $Q=M$ in the above lemma, we get a contradiction. ∎

In the following three lemma and corollaries, we suppose that $p$ is a Mersenne prime of the form $2^{m}-1$ (with $m$ prime).

Lemma 3.13**.**

Let $P,Q\in\mathbb{F}_{2}[x]$ such that $P$ is irreducible of degree $m$ and $P\nmid Q(Q+1)$. Then, $P$ divides $Q^{p}+1$.

Proof.

Let $\beta$ be a root of $P$. $P$ and $\beta$ are primitive, $ord(\beta)=2^{m}-1=p$, $Q(\beta)\not\in\{0,1\}$. Thus, $Q(\beta)=\beta^{t}$ for some $1\leq t\leq p-1$. Hence, $Q(\beta)^{p}=\beta^{tp}=1.$ So, $P$ divides $Q^{p}+1$. ∎

Corollary 3.14**.**

Any irreducible polynomial $P\not=M$ $($Mersenne or not$)$, of degree $m$, divides $\sigma(M^{p-1})$.

Proof.

$P$ does not divide $x^{a}(x+1)^{b}M=M(M+1)=Q(Q+1)$. So, we apply Lemma 3.13 to $Q=M$. ∎

Corollary 3.15**.**

*The polynomial $1+x+x^{2}$ divides $\sigma(M^{p-1})$ if and only if $(M\not=1+x+x^{2}$ and $p=3)$,

$1+x^{2}+x^{3}$ divides $\sigma(M^{p-1})$ if and only if $M\not=1+x^{2}+x^{3}$ and $p=7$,

$1+x+x^{3}$ divides $\sigma(M^{p-1})$ if and only if $M\not=1+x+x^{3}$ and $p=7$.*

Proof.

Apply Corollary 3.14 with $m\in\{2,3\}$. ∎

3.2 Case $M\in\{M_{1},M_{3},\overline{M_{3}}\}$

Lemma 3.2 implies that $M\not=M_{1}$. It suffices to suppose that $M=M_{3}$.

We refer to Section 5.2 in [20]. Put $U:=M_{1}M_{2}\overline{M_{2}}$. By [20, Lemma 5.4], we have to consider four cases:

i) $\gcd(\sigma(M^{2h}),U)=1$,

ii) $\sigma(M^{2h})=M_{1}B$, with $\gcd(B,U)=1$,

iii) $\sigma(M^{2h})=M_{2}\overline{M_{2}}B$, with $\gcd(B,U)=1$,

iv) $\sigma(M^{2h})=UB$, with $\gcd(B,U)=1$,

where any irreducible divisor of $B$ has degree exceeding $5$.

We get Lemma below which contradicts the fact that $U_{2h}$ is a square.

Lemma 3.16**.**

$\alpha_{3}(U_{2h})=1$* or $\alpha_{5}(U_{2h})=1$.*

Proof.

For i), iii) and iv) : use Lemmas 5.9, 5.10, 5.15, 5.17 (still in [20]).

For ii): since $\sigma(M^{2h})=(x^{2}+x+1)B$ and $U_{2h}=(x^{2}+x)\sigma(B)$, we obtain (by Lemmas 3.4 and 3.6):

[TABLE]

Thus, $\alpha_{3}(U_{2h})=\alpha_{3}(B)+\alpha_{2}(B)=\alpha_{1}(B)=1$. ∎

3.3 Case where $M\in\{M_{2},\overline{M_{2}}\}$ and $h\geq 2$

It suffices to consider $M=M_{2}=1+x+x^{3}$. Recall that $U_{2h}=\sigma(\sigma(M^{2h}))$ splits and it is a square. Note also that if $h=1$, then $\sigma({M_{2}}^{2h})=\sigma({M_{2}}^{2})=M_{1}M_{3}$.

For $h\in\{2,3\}$, we get by direct computations, $U_{4}=x^{3}(x+1)^{6}(x^{3}+x+1)$ and $U_{6}=x^{8}(x+1)^{4}(x^{3}+x+1)^{2}$ which do not split (even if $U_{6}$ is a square).

So, $h\geq 4$.

Lemma 3.17**.**

*i) $1+x+x^{2}$ divides $\sigma(M^{2h})$ if and only if $3$ divides $2h+1$.

ii) $1+x^{2}+x^{3}$ divides $\sigma(M^{2h})$ if and only if $7$ divides $2h+1$.

iii) Any irreducible divisor of $\sigma(M^{2h})$ is of degree at least $4$, if $2h+1$ is divisible by a prime $p\not\in\{3,7\}$.*

Proof.

i) and ii): from Corollaries 3.12 and 3.14.

iii) follows from i) and ii). ∎

3.3.1 Case where $2h+1$ is divisible by a prime $p\not\in\{3,7\}$

By Lemma 3.7, $\sigma(M^{p-1})$ divides $\sigma(M^{2h})$. So, we may suppose that $2h+1=p$ so that $2h=p-1$. It suffices then to prove (directly or by a contradiction) that $\sigma(M^{2h})$ is divisible by a non Mersenne prime.

Lemma 3.18**.**

*i) $\alpha_{l}(U_{2h})=\alpha_{l}(\sigma(M^{2h}))$ for $l\in\{1,2,3\}$.

ii) $\alpha_{l}(\sigma(M^{2h}))=\alpha_{l}(M^{2h})$ for $l\in\{1,2\}$, $\alpha_{3}(\sigma(M^{2h}))=\alpha_{3}(M^{2h}+M^{2h-1})$.*

Proof.

i) follows from Lemma 3.17.

ii): for $l\leq 2$, one has: $6h-l=\deg(\sigma(M^{2h}))-l=\deg((M^{2h})-l>3(2h-1)=\deg(M^{2h-1})$ and for $3\leq l\leq 5$, $6h-l>3(2h-2)=\deg(M^{2h-2})$. Hence, we get ii). ∎

Corollary 3.19**.**

$\alpha_{3}(U_{2h})=1$* if $h\geq 4$.*

Proof.

$\alpha_{3}(U_{2h})=\alpha_{3}(M^{2h}+M^{2h-1})=\alpha_{3}[(x^{3}+x)M^{2h-1}]=\alpha_{3}(M^{2h-1})+\alpha_{1}(M^{2h-1})$. But, $M^{2h-1}=(x^{3}+x+1)^{2h-1}=(x^{3}+x)^{2h-1}+(x^{3}+x)^{2h-2}+\cdots$

So, $\alpha_{3}(M^{2h-1})$ (resp. $\alpha_{1}(M^{2h-1})$) which is the coefficient of $x^{6h-6}$ (resp. of $x^{6h-4}$) in $M^{2h-1}$, equals $1$ (resp. [math]). ∎

3.3.2 Case where $7$ divides $2h+1$

In this case, by Lemma 3.7, $\sigma(M^{6})$ divides $\sigma(M^{2h})$, where $\sigma(M^{6})=(x^{3}+x^{2}+1)(x^{6}+x^{5}+1)(x^{9}+x^{7}+x^{5}+x+1)$ is divisible by the non Mersenne prime $x^{9}+x^{7}+x^{5}+x+1=1+x(x+1)^{2}(x^{3}+x+1)^{2}$.

3.3.3 Case where $3$ is the unique prime factor of $2h+1$

In this case, $2h+1=3^{w}$, with $w\geq 2$ because $2h+1\geq 9$. So, $9$ divides $2h+1$ and thus $\sigma(M^{8})$ divides $\sigma(M^{2h})$ (by Lemma 3.7). We are done because $\sigma(M^{8})=(x^{2}+x+1)(x^{4}+x^{3}+1)(x^{6}+x+1)(x^{12}+x^{8}+x^{7}+x^{4}+1)$, where $x^{6}+x+1=1+x(x+1)M_{3}$ is not a Mersenne prime.

3.4 Case where $M\not\in{\mathcal{M}}$ and $2h+1$ is divisible by a Mersenne prime number $p\not=7$

Set $p:=2^{m}-1$, where $m$ and $p$ are both prime. We shall prove that $\sigma(M^{p-1})$ is divisible by a non Mersenne prime. Note that there are (at present) “only” 51 known Mersenne prime numbers (OEIS Sequences A$000043$ and A$000668$). The first five of them are: $3,7,31,127$ and $8191$.

Here, $a+b=\deg(M)\geq 5$ since $M\not\in{\mathcal{M}}$. Corollary 3.14 and Lemma 3.1-iii) imply that for $p\geq 31$, we get our result. It remains then the case $p=3$ because $p\not=7$.

Lemma 2.2 has already treated the case where $\omega(\sigma(M^{2}))=2$. So, we suppose that $\omega(\sigma(M^{2}))\geq 3$. Put:

[TABLE]

We get by Corollary 3.15:

Lemma 3.20**.**

*i) $1+x+x^{2}$ divides $\sigma(M^{2})$.

ii) No irreducible polynomial of degree $r\geq 3$ such that $2^{r}-1$ is prime divides $\sigma(M^{2})$.*

Lemma 3.21**.**

Write $\sigma(M^{2})=M_{1}B$ where $M_{1}=1+x+x^{2}$, $\gcd(M_{1},B)=1$. One has:

[TABLE]

Proof.

$\sigma(M^{2})=M_{1}B=(x^{2}+x+1)B$. So we directly get i) and ii).

iii): $\sigma(M^{2})=1+M+M^{2}=x^{2a}(x+1)^{2b}+x^{a}(x+1)^{b}+1$.

$2a+2b-3>a+b$ because $a+b\geq 4$ and $x^{2a}(x+1)^{2b}$ is a square. So, $\alpha_{3}(\sigma(M^{2}))=\alpha_{3}(x^{2a}(x+1)^{2b})=0$.

∎

Lemma 3.22**.**

One has:

[TABLE]

Proof.

$W=\sigma(\sigma(M^{2}))=\sigma(M_{1}B)=\sigma(M_{1})\sigma(B)=(x^{2}+x)\sigma(B)$.

Moreover, any irreducible divisor of $B$ has degree more than $3$. Hence, $\alpha_{l}(\sigma(B))=\alpha_{l}(B),$ for $1\leq l\leq 3$. One gets:

[TABLE]

and $\alpha_{3}(W)=\alpha_{3}(\sigma(B))+\alpha_{2}(\sigma(B))=\alpha_{3}(B)+\alpha_{2}(B)$. ∎

Corollary below contradicts the fact that $W$ is a square and finishes the proof for $p=3$.

Corollary 3.23**.**

$\alpha_{3}(W)=1$.

Proof.

$W$ is a square, so $0=\alpha_{1}(W)=\alpha_{1}(B)+1$ and thus $\alpha_{1}(B)=1$.

Lemma 3.21-iii) implies that $0=\alpha_{3}(\sigma(M^{2}))=\alpha_{3}(B)+\alpha_{2}(B)+\alpha_{1}(B)$. Therefore, we get: $\alpha_{3}(W)=\alpha_{3}(B)+\alpha_{2}(B)=\alpha_{1}(B)=1.$ ∎

Remark 3.24**.**

Our method fails for $p=7$. Indeed, for many $M$’s, one has $\alpha_{3}(W)=\alpha_{5}(W)=0$ so that we do not reach a contradiction. We should find a large enough odd integer $l$ such that $\alpha_{l}(W)=0$. But, this does not appear always possible.**

3.5 Case where $M\not\in{\mathcal{M}}$ and $2h+1$ is divisible by a prime $p$ with $ord_{p}(2)\equiv 0\mod 8$

Lemmas 3.25 and 3.8 imply Corollary 3.26.

Lemma 3.25**.**

There exists no Mersenne prime of degree multiple of $8$.

Proof.

If $Q=1+x^{c_{1}}(x+1)^{c_{2}}$ with $c_{1}+c_{2}=8k$, then $\omega(Q)$ is even by [22, Corollary 3.3]. ∎

Corollary 3.26**.**

If $2h+1$ is divisible by a prime $p$ such that $8$ divides $ord_{p}(2)$ $($in particular, when $p>5$ is a Fermat prime$)$, then $\sigma(M^{2h})$ is divisible by a non Mersenne prime.

Proof.

If not, Lemma 3.8 implies that $ord_{p}(2)$ divides $\deg(P_{j})$, for any $j\in J$. So, we get a contradiction to Lemma 3.25: $8$ divides $\deg(P_{j})$.

In particular, if $p=2^{2^{w}}+1$, with $w\geq 2$, then $ord_{p}(2)=2^{w+1}$ which is divisible by $8$. ∎

Remarks 3.27**.**

i) If $p$ is a Fermat prime, then $ord_{p}(2)\equiv 0\mod 8$. The converse is false. Examples: $p\in\{97,673\}$ with $ord_{p}(2)=48$.

*ii) It remains the following (large) case to complete the proof of Conjecture 2.1: $M\not\in\mathcal{M}$ and $2h+1$ is divisible by $p\in\{5,7\}$ or by $p>7$ which is neither Mersenne prime nor Fermat prime.

Moreover, assuming Conjecture 2.1, similar proofs as in Section 2.1.1 would state that there exists no odd perfect polynomial over $\mathbb{F}_{2}$ which is only divisible by Mersenne primes.***