Bi-Unitary Superperfect Polynomials over 𝔽₂ with at Most Two Irreducible Factors

Abstract

A divisor B of a nonzero polynomial A, defined over the prime field of two elements, is unitary (resp. bi-unitary) if $gcd(B,A/B)=1$ (resp. $gc{d}_u(B,A/B)=1)$, where $gc{d}_u(B,A/B)$ denotes the greatest common unitary divisor of B and $A/B$. We denote by ${\sigma }^{**}\left(A\right)$ the sum of all bi-unitary monic divisors of $A.$ A polynomial A is called a bi-unitary superperfect polynomial over ${\mathbb{F}}_2$ if the sum of all bi-unitary monic divisors of ${\sigma }^{**}\left(A\right)$ equals $A.$ In this paper, we give all bi-unitary superperfect polynomials divisible by one or two irreducible polynomials over ${\mathbb{F}}_2$. We prove the nonexistence of odd bi-unitary superperfect polynomials over ${\mathbb{F}}_2$.

1. Introduction

Let n and k be positive integers, and let $\sigma \left(n\right)$ (resp. ${\sigma }^{*}\left(n\right)$) denote the sum of positive (resp. unitary) divisors of the integer n. A divisor d of n is unitary if d and $n/d$ are coprime. We call the number n a k-superperfect number if ${\sigma }^k\left(n\right)=\underset{k\text{-}\mathrm{times}}{\underbrace{\sigma \left(\sigma \right(...\left(\sigma \right(}}n\left)\right)\left)\right)=2n$. When $k=1$, n is called a perfect number. An integer $M=2^p-1,$ where p is a prime number, is called a Mersenne number. It is also well known that an even integer n is perfect if and only if $n=M(M+1)/2$ for some Mersenne prime number M. Suryanarayana [1] considered k-superperfect numbers in the case $k=2$. Numbers of the form $2^{p-1}$ (p is prime) are 2-superperfect if $2^{p-1}-1$ is a Mersenne prime. It is not known if there are odd k-superperfect numbers. Sitaramaiah and Subbarao [2] studied the unitary superperfect numbers, with the integers n satisfying ${\sigma }^{*2}\left(n\right)={\sigma }^{*}\left({\sigma }^{*}\left(n\right)\right)=2n.$ They found all unitary superperfect numbers below ${10}^8.$ The first unitary superperfect numbers are $2,9,165,$ and 238. A positive integer n has a bi-unitary divisor, d, if the greatest common unitary divisor of d and $n/d$ is equal to 1. The arithmetic function ${\sigma }^{**}\left(n\right)$ denotes the sum of positive bi-unitary divisors of the integer n. Wall [3] proved that there are only three bi-unitary perfect numbers $({\sigma }^{**}\left(n\right)=2n)$, namely, 6, 60, and 90. Yamada [4] proved that 2 and 9 are the only bi-unitary superperfect numbers, that is, ${\sigma }^{**2}\left(n\right)=2n$ if and only if $n\in \{2,9\}$.

Here, let A be a nonzero polynomial over the prime field ${\mathbb{F}}_2$. We say that A is a splitting polynomial if it can be factored completely into linear factors over ${\mathbb{F}}_2$. A divisor B of A is unitary (resp. bi-unitary) if $gcd(B,A/B)=1$ (resp. $gc{d}_u(B,A/B)=1)$, where $gc{d}_u(A,A/B)$ denotes the greatest common unitary divisor of B and $A/B$. We denote by $\sigma$ the sum of the monic divisors B of $A,$ that is, $\sigma \left(A\right)={\sum }_{B\mid A}B.$${\sigma }^{*}\left(A\right)$ (resp. ${\sigma }^{**}\left(A\right))$ represents the sum of all unitary (resp. bi-unitary) monic divisors of $A.$ Note that all the functions $\sigma$, ${\sigma }^{*}$, and ${\sigma }^{**}$ are multiplicative and degree-preserving.

We say that a polynomial A is an even polynomial if it has a linear factor in ${\mathbb{F}}_2\left[x\right]$; otherwise, it is an odd polynomial. A polynomial M of the form $1+x^a{(x+1)}^b$ is called Mersenne. The first five Mersenne polynomials over ${\mathbb{F}}_2$ are $M_1=1+x+x^2$, $M_2=1+x+x^3$, $M_3=1+x^2+x^3$, $M_4=1+x+x^2+x^3+x^4$, and $M_5=1+x^3+x^4.$ Note that all these polynomials are irreducible, so we call them Mersenne primes.

Notations: We use the following notations throughout the article:

• $\mathbb{N}$ (resp. ${\mathbb{N}}^{*})$ represents the set of non-negative (resp. positive) integers.
• deg$\left(A\right)$ denotes the degree of the polynomial $A.$
• $\overline{A}$ is the polynomial obtained from A with x replaced by $x+1$, that is, $\overline{A}\left(x\right)=A(x+1)$.
• P and Q are distinct irreducible non-constant polynomials.
• $P_i$ and $Q_j$ are distinct odd irreducible non-constant polynomials.

Let $\omega \left(A\right)$ denote the number of distinct irreducible monic polynomials that divide A. The notion of a perfect polynomial over ${\mathbb{F}}_2$ was introduced first by Canaday [5]. A polynomial A is perfect if $\sigma \left(A\right)=A.$ Canaday studied the case of even perfect polynomials with $\omega \left(A\right)\le 3.$ In the past few years, Gallardo and Rahavandrainy [6,7,8] showed the non-existence of odd perfect polynomials over ${\mathbb{F}}_2$ with either $\omega \left(A\right)=3$ or with $\omega \left(A\right)\le 9$ in the case where all exponents of the irreducible factors of A are equal to 2. A polynomial A is said to be a unitary (resp. a bi-unitary) perfect if ${\sigma }^{*}\left(A\right)=A$ (resp. ${\sigma }^{**}\left(A\right)=A$). Furthermore, A is called a unitary (resp. a bi-unitary) superperfect if ${\sigma }^{*2}\left(A\right)={\sigma }^{*}\left({\sigma }^{*}\left(A\right)\right)=A$ (resp. ${\sigma }^{**2}\left(A\right)={\sigma }^{**}\left({\sigma }^{**}\left(A\right)\right)=A)$.

Note that the function ${\sigma }^{**2}$ is degree-preserving but not multiplicative, and this is the main challenge in this work. Thus, working on bi-unitary superperfect polynomials over ${\mathbb{F}}_2$ is not an easy task especially when A is divisible by more than two irreducible factors.

In this paper, we prove the non-existence of odd bi-unitary superperfect polynomials A when A is divisible by at least two irreducible factors (Corollary 4). We give a complete classification for all bi-unitary superperfect polynomials over ${\mathbb{F}}_2$ that are divisible by at most two distinct irreducible factors (Theorems 1 and 2). Bi-unitary superperfect polynomials over ${\mathbb{F}}_2$ that are neither unitary perfect nor bi-unitary perfect are found. The polynomials $x^4{(x+1)}^4,x^9{(x+1)}^9,x^9{(x+1)}^{13}$, and $x^2{(x+1)}^{2^d-1}$, d is a positive integer, are examples of bi-unitary superperfect polynomials that are neither unitary perfect nor bi-unitary perfect.

Our main results are given in the following theorems:

Theorem 1. 

Let A be a bi-unitary superperfect over ${\mathbb{F}}_2$ such that $\omega \left(A\right)=1$; then, $A,\overline{A}\in \{x^2,x^{2^d-1}\},$ where $d\in {\mathbb{N}}^{*}.$

Theorem 2. 

Let A be a bi-unitary superperfect over ${\mathbb{F}}_2$ such that $\omega \left(A\right)=2$; then, $A,\overline{A}\in \{x^2{(x+1)}^2,x^4{(x+1)}^4,x^9{(x+1)}^9,x^9{(x+1)}^{13},x^2{(x+1)}^{2^d-1},x^{2^{d_1}-1}{(x+1)}^{2^{d_2}-1}\},$ where $d,d_1,d_2\in {\mathbb{N}}^{*}.$

2. Previous Work

Many researchers studied the unitary perfect polynomials over ${\mathbb{F}}_2$. In their works [7,8], the authors listed the unitary perfect polynomials over ${\mathbb{F}}_2$, where $\omega \left(A\right)$ does not exceed 4. They listed others that are divisible by $x(x+1)M,$ where M is a Mersenne polynomial, raised to certain powers. They proved that the only unitary perfect polynomials over ${\mathbb{F}}_2$ of the form $A=x^a{(x+1)}^b{\prod }_{i=1}{M}_i$ and $h_i=2^{n_i}$, $n_i\in {\mathbb{N}}^{*}$ are those of the form $B^{2n}$ or ${\overline{B}}^{2n}$, where

$$B\in \left\{\begin{array}{cc}{x}^3{(x+1)}^3{M}_1^2,x^3{(x+1)}^2{M}_1,x^5{(x+1)}^4{M}_4\hfill & \hfill \mathrm{if}\omega \left(A\right)\le 3,\\ x^7{(x+1)}^4{M}_2{M}_3,x^5{(x+1)}^6{M}_1^2{M}_4,x^5{(x+1)}^5{M}_4{M}_5,x^7{(x+1)}^7{M}_2^2{M}_3^2\hfill & \hfill \mathrm{if}\omega \left(A\right)=4,\\ x^7{(x+1)}^6{M}_2^1{M}_2{M}_3,x^7{(x+1)}^5{M}_2{M}_3{M}_5\hfill & \hfill \mathrm{if}\omega \left(A\right)=5.\end{array}\right.$$

In [9], Beard found many bi-unitary perfect polynomials over ${\mathbb{F}}_{p^d}$, some of which are neither perfect nor unitary perfect. Beard showed that the only bi-unitary perfect polynomials over ${\mathbb{F}}_2$ with exactly two prime factors are $x^2{(x+1)}^2$ and $x^{2^{n-1}}{(x+1)}^{2^{n-1}}$, for any $n\in {\mathbb{N}}^{*}$ (Theorem 5 in [9]). He conjectured a characterization of the bi-unitary perfect polynomials, which splits over ${\mathbb{F}}_p$ when $p>2.$ Beard also gave examples of non-splitting bi-unitary perfect polynomials over ${\mathbb{F}}_p$ when $p\in \{2,3,5\}$. Rahavandrainy [10] gave all bi-unitary perfect polynomials over the prime field ${\mathbb{F}}_2$, with at most four irreducible factors (Lemmas 7 and 8).

Gallardo and Rahavandrainy [11] classified some unitary superperfect polynomials with a small number of prime divisors under some conditions on the number of prime factors of ${\sigma }^{*}\left(A\right)$. They proved that $A\in {\mathbb{F}}_2\left[x\right]$ is a unitary superperfect polynomial if

$$A\in \left\{\begin{array}{cc}{x}^{2^n}{(x+1)}^{2^m},x^{3·2^n}{(x+1)}^{3·2^m},x^3{(x+1)}^5,x{(x+1)}^5,x^7{(x+1)}^7\hfill & \hfill \mathrm{if}\omega \left(A\right)=2,\\ x^2{(x+1)}^3{M}_1,x^3{(x+1)}^3{M}_1^a,x{(x+1)}^5{M}_1^a,x{(x+1)}^5(x^3+x^2+1)\hfill & \hfill \mathrm{if}\omega \left(A\right)=3.\end{array}\right.$$

For some m, $n\in {\mathbb{N}}^{*}$ and $a\in \{1,2\}$.

3. Preliminaries

The following two lemmas are helpful.

Lemma 1. 

Let A be a polynomial in ${\mathbb{F}}_2\left[x\right]$; then, ${\sigma }^{*}\left(A^{2^n}\right)={\left({\sigma }^{*}\left(A\right)\right)}^{2^n}$ and n is a non-negative integer.

Proof. 

The result follows since ${\sigma }^{*}$ is multiplicative and ${\sigma }^{*}\left(p^{2^n}\right)=1+p^{2^n}={(1+p)}^{2^n}={\left({\sigma }^{*}\left(p\right)\right)}^{2^n}$. □

Lemma 2. 

If A is a unitary superperfect polynomial over ${\mathbb{F}}_2$, then $A^{2^n}$ is also a unitary superperfect polynomial over ${\mathbb{F}}_2$ for all non-negative integers n.

Proof. 

Let A be a unitary superperfect, and let $B={\sigma }^{*}\left(A\right)$. By Lemma 1, we have ${\sigma }^{*2}\left(A^{2^n}\right)={\sigma }^{*}\left({\sigma }^{*}\left(A^{2^n}\right)\right)={\sigma }^{*}\left(B^{2^n}\right)={\left({\sigma }^{*}\left(B\right)\right)}^{2^n}={\left({\sigma }^{*}\left({\sigma }^{*}\left(A\right)\right)\right)}^{2^n}=A^{2^n}$. □

Lemma 3 

(Lemma 2.4 in [11]). Let A be a polynomial in ${\mathbb{F}}_2\left[x\right].$

(1) 

If P is an odd prime factor of $A,$ then $x(x+1)$ divides ${\sigma }^{*}\left(A\right).$

(2) 

If $x(x+1)$ divides $A,$ then $x(x+1)$ divides ${\sigma }^{*}\left(A\right).$

(3) 

If A is unitary superperfect that has an odd prime factor, then $x(x+1)$ divides $A.$

The following results are needed, and they are a result of Beard’s [9] and Rahavandrainy’s [10] works.

Lemma 4 

(Theorem 1 and its Corollary in [9]). If A is a non-constant bi-unitary perfect polynomial, then $x(x+1)$ divides A and $\omega \left(A\right)$$\ge 2.$

Proposition 1 

(Lemma 2.2 in [10]).

(1) 

${\sigma }^{**}\left(P^{2a+1}\right)=\sigma \left(P^{2a+1}\right).$

(2) 

${\sigma }^{**}\left(P^{2a}\right)=(1+P^{a+1})\sigma \left(P^{a-1}\right)=(1+P)\sigma \left(P^a\right)\sigma \left(P^{a-1}\right).$

The table in Section 7 shows some values of ${\sigma }^{**}\left(A\right)$ when A is a power of the first five Merssene primes.

Corollary 1. 

If a is a positive integer, then

(1) 

1+x divides ${\sigma }^{**}\left(x^a\right)$.

(2) 

x divides ${\sigma }^{**}\left({(1+x)}^a\right)$.

Proof. 

An immediate result of Proposition 1. □

Corollary 2 

(Corollary 2.3 in [10]). Let $T\in {\mathbb{F}}_2\left[x\right]$ be irreducible. Then,

(i) 

If $a\in \{4r,4r+2\}$, where $2r-1$ or $2r+1$ is of the form $2^{\alpha }u-1$, u odd, then ${\sigma }^{**}\left(P^a\right)={(1+P)}^{2^{\alpha }}·\sigma \left(P^{2r}\right)·{\left(\sigma \left(P^{u-1}\right)\right)}^{2^{\alpha }},gcd(\sigma \left(P^{2r}\right),\sigma \left(P^{u-1}\right))=1.$

(ii) 

If $a=2^{\alpha }u-1$ is odd, with u odd, then ${\sigma }^{**}\left(P^a\right)={(1+P)}^{2^{\alpha }-1}·{\left(\sigma \left(P^{u-1}\right)\right)}^{2^{\alpha }}$.

The proof of the below proposition follows from Proposition 1 and the binomial formula.

Proposition 2. 

Let the polynomial $M_i$ be the Mersenne prime and $Q_j$ be an irreducible polynomial over ${\mathbb{F}}_2$, and let $a,c\in {\mathbb{N}}^{*}.$ If ${\alpha }_j\in \mathbb{N},$ then

(1) 

$x(x+1)$ divides ${\sigma }^{**}\left(M_i^c\right).$

(2) 

${\sigma }^{**}\left(M_1^c\right)=x^a{(x+1)}^a\underset{j}{\Pi }{Q}_j^{{\alpha }_j}.$

(3) 

${\sigma }^{**}\left(M_2^c\right)=x^a{(x+1)}^{2a}\underset{j}{\Pi }{Q}_j^{{\alpha }_j}.$

(4) 

${\sigma }^{**}\left(M_3^c\right)=x^{2a}{(x+1)}^a\underset{j}{\Pi }{Q}_j^{{\alpha }_j}.$

(5) 

${\sigma }^{**}\left(M_4^c\right)=x^a{(x+1)}^{3a}\underset{j}{\Pi }{Q}_j^{{\alpha }_j}.$

(6) 

${\sigma }^{**}\left(M_5^c\right)=x^{3a}{(x+1)}^a\underset{j}{\Pi }{Q}_j^{{\alpha }_j}.$

Proposition 3 

(Corollary 2.4 in [10]).

(1) 

${\sigma }^{**}\left(x^a\right)$ splits over ${\mathbb{F}}_2$ if and only if $a=2$ or $a=2^d-1$, for some $d\in {\mathbb{N}}^{*}$.

(2) 

${\sigma }^{**}\left(P^c\right)$ splits over ${\mathbb{F}}_2$ if and only if P is Mersenne and $c=2$ or $c=2^d-1$ for some $d\in {\mathbb{N}}^{*}$.

Lemma 5 summarizes key results taken from Canaday’s paper [5].

Lemma 5. 

Let T be irreducible in ${\mathbb{F}}_2\left[x\right]$ and let $n,m\in \mathbb{N}$.

(i) 

If T is a Mersenne prime and if $T=T^{*}$, then $T\in \{M_1,M_4\}$.

(ii) 

If $\sigma \left(x^{2n}\right)=PQ$ and $P=\sigma \left({(x+1)}^{2m}\right)$, then $2n=8$, $2m=2$, $P=M_1$, and $Q=P\left(x^3\right)=1+x^3+x^6$.

(iii) 

If any irreducible factor of $\sigma \left(x^{2n}\right)$ is a Mersenne prime, then $2n\le 6$.

(iv) 

If $\sigma \left(x^{2n}\right)$ is a Mersenne prime, then $2n\in \{2,4\}$.

Lemma 6 

(Lemma 2.6 in [12]). Let $m\in {\mathbb{N}}^{*}$ and M be a Mersenne prime. Then, $\sigma \left(x^{2m}\right)$, $\sigma \left({(x+1)}^{2m}\right)$, and $\sigma \left(M^{2m}\right)$ are all odd and square-free.

4. Bi-Unitary Superperfect Polynomials

Recall that A is a bi-unitary superperfect polynomial in ${\mathbb{F}}_2\left[x\right]$ if ${\sigma }^{**2}\left(A\right)={\sigma }^{**}\left({\sigma }^{**}\left(A\right)\right)=A.$ The polynomial $A=x^4{(1+x)}^4$ is a bi-unitary superperfect polynomial over ${\mathbb{F}}_2$.

The following polynomials are considered over ${\mathbb{F}}_2:$

$\begin{array}{ccc}C=1+x+x^4,\hfill & B_1=x^3{(x+1)}^4{M}_1,\hfill & B_2=x^3{(x+1)}^5{M}_1^2,\hfill \\ B_3=x^4{(x+1)}^4{M}_1^2,\hfill & B_4=x^6{(x+1)}^6{M}_1^2,\hfill & B_5=x^4{(x+1)}^5{M}_1^3,\hfill \\ B_6=x^7{(x+1)}^8{M}_5,\hfill & B_7=x^7{(x+1)}^9{M}_5^2,\hfill & B_8=x^8{(x+1)}^8{M}_4{M}_5,\hfill \\ B_9=x^8{(x+1)}^9{M}_4{M}_5^2,\hfill & B_{10}=x^7{(x+1)}^{10}{M}_1^2{M}_5,\hfill & B_{11}=x^7{(x+1)}^{13}{M}_2^2{M}_3^2,\hfill \\ B_{12}=x^9{(x+1)}^9{M}_4^2{M}_5^2,\hfill & B_{13}=x^{14}{(x+1)}^{14}{M}_2^2{M}_3^2,\hfill & R_1=x^4{(x+1)}^5{M}_1^{4}C,\hfill \\ R_2=x^4{(x+1)}^5{M}_1^5{C}^2.\hfill \end{array}$

The proof of the following lemmas follow directly.

Proposition 4. 

If A is a bi-unitary perfect polynomial over ${\mathbb{F}}_2$, then A is also a bi-unitary superperfect polynomial.

Proposition 5. 

If A is a bi-unitary superperfect polynomial over ${\mathbb{F}}_2$, then $B={\sigma }^{**}\left(A\right)$ is also a bi-unitary superperfect polynomial.

Rahavandrainy (Lemma 2.6 in [10]) proved that if A is a bi-unitary perfect polynomial over ${\mathbb{F}}_2$, where $A=A_1{A}_2$ such that $gcd(A_1,A_2)=1$, then $A_1$ is a bi-unitary perfect polynomial if and only if $A_2$ is a bi-unitary perfect polynomial. Rahavandrainy’s previous result is not valid in the case of bi-unitary superperfect polynomials because the bi-unitary superperfect polynomial $A=x^2{(1+x)}^2{(1+x+x^2)}^2$ is a counterexample over ${\mathbb{F}}_2$. In fact, $A_1=x^2{(1+x)}^2$ is a bi-unitary superperfect, but $A_2={(1+x+x^2)}^2$ is not a bi-unitary superperfect.

Lemma 7 

(Theorem 1.1 in [10]). Let $A\in {\mathbb{F}}_2\left[x\right]$ be a bi-unitary perfect polynomial such that $\omega \left(A\right)=3$. Then, $A,\overline{A}\in \{B_j:j\le 7\}.$

Lemma 8 

(Theorem 1.2 in [10]). Let $A\in {\mathbb{F}}_2\left[x\right]$ be a bi-unitary perfect polynomial such that $\omega \left(A\right)=4$. Then $A,\overline{A}\in \{B_j:8\le j\le 13\}\bigcup \{R_1,R_2\}.$

Proposition 6. 

If $A\left(x\right)$ is a bi-unitary superperfect polynomial over ${\mathbb{F}}_2$, then so is $\overline{A}\left(x\right)$.

Lemma 9. 

$x(x+1)$ divides ${\sigma }^{**}\left(P^a\right)$, a is a positive integer.

Proof. 

Since P is odd, then $P\left(0\right)=P\left(1\right)=1$. If $a=2n+1$, then ${\sigma }^{**}\left(P^{2n+1}\right)\left(0\right)=1+\underset{(2n+1)\text{-}\mathrm{times}}{\underbrace{P\left(0\right)+\dots +P^{2n+1}\left(0\right)}}=1+2n+1=0$. If $a=2n$, then $1+P^{n+1}\left(0\right)=0.$ Thus, x divides ${\sigma }^{**}\left(P^a\right)$ for every a $\in \mathbb{N}$. Similarly, $x+1$ divides ${\sigma }^{**}\left(P^a\right)$. Hence, $x(x+1)$ divides ${\sigma }^{**}\left(P^a\right)$. □

Lemma 10. 

Let A be a polynomial in ${\mathbb{F}}_2\left[x\right]$.

(1) 

If P is an odd prime factor of $A,$ then $x(x+1)$ divides ${\sigma }^{**}\left(A\right).$

(2) 

If $x(x+1)$ divides $A,$ then $x(x+1)$ divides ${\sigma }^{**}\left(A\right).$

Proof. 

(1)

We write $A=P^{a}B$, where $a\in {\mathbb{N}}^{*}$ and $B\in {\mathbb{F}}_2\left[x\right]$ such that $gcd(P,B)=1.$ However, $1+P$ divides ${\sigma }^{**}\left(A\right)$, and the result follows since $x(x+1)$ divides $1+P$.

(2)

In a similar manner, we write $A=x^a{(x+1)}^{b}B$, where $a,b\in {\mathbb{N}}^{*}$.

□

Corollary 3. 

If $A\in {\mathbb{F}}_2\left[x\right]$ and $\omega \left(A\right)$$\ge 2$, then $x(x+1)$ divides ${\sigma }^{**}\left(A\right).$

Proof. 

Let $\omega \left(A\right)$$\ge 2$. If $x(x+1)$ divides A, then Corollary 1 is completed. If $x(x+1)$ does not divide A, then A is divisible by an irreducible polynomial $P\notin${$x,1+x\}$, and the result follows using Lemma 9. □

Corollary 4. 

Let A be a polynomial in ${\mathbb{F}}_2\left[x\right]$ with $\omega \left(A\right)\ge 2$. If A is a bi-unitary superperfect, then $x(x+1)$ divides $A.$

Proof. 

Let $A={\sigma }^{**2}\left(A\right)={\sigma }^{**}\left(B\right),\mathrm{where}B={\sigma }^{**}\left(A\right)$. Since $\omega \left(A\right)\ge 2$, then either P or $x(x+1)$ divides A. In both cases, $x(x+1)$ divides ${\sigma }^{**}\left(A\right)=B$ (Lemma 10). Thus, $x(x+1)$ divides ${\sigma }^{**}\left(B\right)={\sigma }^{**2}\left(A\right)$. □

The below corollary follows directly from Corollary 4.

Corollary 5. 

If $A=P^a{Q}^b$ and $a,b\in {\mathbb{N}}^{*}.$ is a bi-unitary superperfect polynomial over ${\mathbb{F}}_2$, then $A=x^a{(x+1)}^b$.

The following lemma is similar to Proposition 3.

Lemma 11. 

Let $a,b\in {\mathbb{N}}^{*}$, then

(1) 

If a is even; then, ${\sigma }^{**2}\left(x^a\right)$ and ${\sigma }^{**2}\left({(x+1)}^a\right)$ splits over ${\mathbb{F}}_2$ if and only if $a\in \left\{2,4,10,12\right\}.$

(2) 

If a is odd, then ${\sigma }^{**2}\left(x^a\right)$ and ${\sigma }^{**2}\left({(x+1)}^a\right)$ splits over ${\mathbb{F}}_2$ if and only if $a\in \left\{5,9,13,2^d-1\right\}$ for some $d\in {\mathbb{N}}^{*}$.

Proof. 

(1)

If ${\sigma }^{**}\left(x^a\right)$ splits, $a=2$ (Proposition 3) and ${\sigma }^{**2}\left(x^a\right)={(x+1)}^2$. Suppose that ${\sigma }^{**}\left(x^a\right)$ does not split with $a=4r,2r-1=2^{\alpha }u-1$, (resp. $a=4r+2,2r+1=2^{\alpha }u-1$), u is odd, $r\ge 1$. However, ${\sigma }^{**2}\left(x^a\right)={\sigma }^{**}\left({(1+x)}^{2^{\alpha }}·\sigma \left(x^{2r}\right)·{\left(\sigma \left(x^{u-1}\right)\right)}^{2^{\alpha }}\right)$; thus, ${\sigma }^{**}\left({(1+x)}^{2^{\alpha }}\right)$ must split. Hence, $\alpha =1$, and since $\sigma \left(x^{2r}\right)$ is odd and square-free (Lemma 6), then $\sigma \left(x^{2r}\right)$ has a Mersenne factor. Thus, $2r\le 6$ and, hence, $u\le 3$.

(2)

Assume $a=2^{\alpha }u-1$, with u is odd. If ${\sigma }^{**}\left(x^a\right)$ splits, then $a=2^d-1$, d is positive (Proposition 3). If ${\sigma }^{**}\left(x^a\right)$ does not split, then $a\ne 2^d-1$ and since ${\sigma }^{**2}\left(x^a\right)=x^{2^{\alpha }-1}·{\sigma }^{**}\left({\left(\sigma \left(x^{u-1}\right)\right)}^{2^{\alpha }}\right)$ splits, $u>1$. Again, using Lemma 6, $\sigma \left(x^{2r}\right)$ has a Mersenne factor. Thus, $u-1\le 6$ and, hence, $u\in \left\{3,5,7\right\}.$ For $u=3$, ${\sigma }^{**2}\left(x^a\right)=x^{2^{\alpha }-1}·{\sigma }^{**}\left({\left(\sigma \left(x^2\right)\right)}^{2^{\alpha }}\right)=x^{2^{\alpha }-1}·{\sigma }^{**}\left(M_1^{2^{\alpha }}\right)$. Hence, $\alpha =1$ and the same result is obtained when $u\in \left\{5,7\right\}$.

The same proof is performed for ${\sigma }^{**2}\left({(x+1)}^a\right)$, and the proof is complete. □

Lemma 12. 

Let a and $b$ have the form $2^n-1$, where $n\in {\mathbb{N}}^{*}$, and let the polynomial $A=1+x^a{(x+1)}^b$ be Mersenne prime over ${\mathbb{F}}_2;$ then, ${\sigma }^{**2}\left(A\right)=x^b{(x+1)}^a.$

Proof. 

Let $a=2^{n_1}-1$ and $b=2^{n_2}-1;$ then,

$$\begin{array}{cc}\hfill {\sigma }^{**2}\left(A\right)& ={\sigma }^{**2}\left(1+x^a{(x+1)}^b\right)\hfill \\ \hfill & ={\sigma }^{**}(\sigma (1+x^a{(x+1)}^b)\hfill \\ \hfill & ={\sigma }^{**}\left(x^a{(x+1)}^b\right)\hfill \\ \hfill & =x^b{(x+1)}^a.\hfill \end{array}$$

□

5. Proof of Theorem 1

We consider the polynomial $A=P^a$ and $a\in {\mathbb{N}}^{*}.$ We prove that ${\sigma }^{**}\left(A\right)$ cannot have more than one prime factor when A is a prime power.

Proposition 7. 

If $A\in \{x,x+1\}$ and ${\sigma }^{**2}\left(A^a\right)$ splits over ${\mathbb{F}}_2$, then A is a bi-unitary superperfect polynomial.

Proof. 

Follows from part (1) of Lemma 11. □

Proposition 8. 

Assume P is odd, then $A=P^{\alpha }\in {\mathbb{F}}_2\left[x\right]$ is not a bi-unitary superperfect polynomial.

Proof. 

Assume $A=P^a$ is a bi-unitary superperfect. Since P divides A, then $x(x+1)$ divides ${\sigma }^{**}\left(A\right)$, and using Lemma 10, we have that $x(x+1)$ divides ${\sigma }^{**2}\left(A\right)=P^a$, a contradiction. □

In particular, if M is a Mersenne prime polynomial over ${\mathbb{F}}_2$, then $M^c$ (c is a positive integer) is never a bi-unitary superperfect polynomial.

Corollary 6. 

Let $a\in {\mathbb{N}}^{*}$ and let $A=P^a$ be a bi-unitary superperfect polynomial over ${\mathbb{F}}_2;$ then, $P\in \left\{x,x+1\right\}$.

It is clear from the preceding two corollaries that a bi-unitary superperfect polynomial must be even.

Lemma 13. 

Let A be a polynomial over ${\mathbb{F}}_2$ with $\omega \left(A\right)=1;$ then, A is a bi-unitary superperfect polynomial if and only if $A,\overline{A}\in \{x^2,x^{2^d-1}\},$ where $d\in {\mathbb{N}}^{*}.$

Proof. 

Using Corollary 6, $A=x^{\alpha }$ or ${\left(x+1\right)}^{\alpha }.$ Assume $A=x^{\alpha }$ and $\alpha =2m$; then, ${\sigma }^{**2}\left(A\right)={\sigma }^{**}\left(\left(x^{m+1}+1\right){\displaystyle \frac{x^m-1}{x-1}}\right).$ Both $x^{m+1}+1$ and $x^m+1$ split over ${\mathbb{F}}_2$ only when $m=1.$ Thus, ${\sigma }^{**2}\left(A\right)={\sigma }^{**}\left(x^2+1\right)=x^2.$ If $\alpha =2m+1$, then ${\sigma }^{**2}\left(A\right)={\sigma }^{**}\left({\displaystyle \frac{x^{2(m+1)}-1}{x-1}}\right)$. The expression $x^{2(m+1)}+1$ splits over ${\mathbb{F}}_2$ when $2m+2=2^d,$$d\in {\mathbb{N}}^{*}$. Then, ${\sigma }^{**2}\left(A\right)={\sigma }^{**}\left({\displaystyle \frac{x^{2^d}-1}{x-1}}\right)=A=x^{2^d-1}.$ The sufficient condition follows via direct computation, and the result follows since if A is a bi-unitary superperfect, then so is $\overline{A}$. □

6. Proof of Theorem 2

We consider the polynomial $A=P^a{Q}^b$ and $a,b\in {\mathbb{N}}^{*}.$ Note that $A=x^2{(1+x)}^2$ and $A=x^{2^{\alpha }-1}{(1+x)}^{2^{\alpha }-1}$ are bi-unitary superperfect polynomials over ${\mathbb{F}}_2$, as shown Proposition 4 and Theorem 5 in [9].

Proposition 9 

(Lemma 3.1 in [10]). If the polynomial ${\sigma }^{**}\left(x^a{(x+1)}^b\right)$ does not split, then $(a\ge 3$ or $b\ge 3)$ and $(a\ne 2^n-1$ or $b\ne 2^m-1$ for any $n,m\ge 1)$.

Lemma 14. 

Let $a,b,d\in {\mathbb{N}}^{*}.$ The polynomial $A=x^a{(x+1)}^b$ is a bi-unitary superperfect over ${\mathbb{F}}_2$ if and only if one of the following is true.

(1) 

If a and b are odd and ${\sigma }^{**}\left(x^a{(x+1)}^b\right)$ splits, then a and b are of the form $2^d-1.$

(2) 

If a and b are odd and ${\sigma }^{**}\left(x^a{(x+1)}^b\right)$ does not split, then $(a,b)\in \left\{\right(9,9),(9,13),(13,9\left)\right\}.$

(3) 

If a and b are even, then $a=b\in \{2,4\}.$

(4) 

If a and b are of opposite parity, then $(a,b)\in \left\{\left(2,2^d-1\right),\left(2^d-1,2\right)\right\}.$

Proof. 

(1)

If $a=2m+1$ and $b=2n+1$, then ${\sigma }^{**2}\left(A\right)={\sigma }^{**}\left({\sigma }^{**}\left(x^a\right){(1+x)}^b\right).$ However, ${\sigma }^{**}\left(x^{2m+1}\right)$ and ${\sigma }^{**}{(x+1)}^{2n+1}$ split over ${\mathbb{F}}_2$ when $2m+1$ and $2n+1$ are of the form $2^d-1$ (Proposition 3).

(2)

If $a=2^{\alpha }u-1$ and $b=2^{\beta }v-1$, $u,v$ are odd. We have $u>1$ and $v>1$ since ${\sigma }^{**}\left(x^a{(x+1)}^b\right)$ does not split. ${\sigma }^{**}\left(x^a{(x+1)}^b\right)={\sigma }^{**}\left({(1+x)}^{2^{\alpha }-1}{\left(\sigma \left(x^{u-1}\right)\right)}^{2^{\alpha }}{x}^{2^{\beta }-1}\sigma {\left({\left(x+1\right)}^{v-1}\right)}^{2^{\beta }}\right).$ Using Proposition 9$(u-1\ge 3$ and $\alpha =1)$ or $(v-1\ge 3$ and $\beta =1)$. Furthermore, $\sigma \left(x^{u-1}\right)$ and $\sigma \left({\left(x+1\right)}^{v-1}\right)$ does not split since ${\sigma }^{**}\left(x^a{(x+1)}^b\right)$ does not split. Thus, there exist Merssene primes M (resp. $M^{\prime })$ that divides $\sigma \left(x^{u-1}\right)$ (resp. $\sigma \left({\left(x+1\right)}^{v-1}\right)$. Hence, $(u-1\le 6)$ or $(v-1\le 6)$, and we have that $u,v\in \left\{5,7\right\}$. If $u=v=5$, then $a=b=9$. If $u=5$ and $v=7$, then $a=9$ and $b=13$. If $u=v=7$, then $a=b=13$ is dismissed.

(3)

If $a,$ b even, then $a\in \{4r,4r+2\}$ such that $2r-1$, $2r+1$ is of the form $2^{\alpha }u-1$, where u is odd and $b\in \{4r^{\prime },4r^{\prime }+2\}$ such that $2r^{\prime }-1$, $2r^{\prime }+1$ is of the form $2^{\beta }$ vs. −1, v odd. Thus,

$${\sigma }^{**}\left(A\right)={(1+x)}^{2^{\alpha }-1}\sigma \left(x^{2r}\right){\left(\sigma \left(x^{u-1}\right)\right)}^{2^{\alpha }}{x}^{2^{\beta }-1}\sigma \left({\left(x+1\right)}^{2r\prime }\right){\left(\sigma \left({\left(x+1\right)}^{v-1}\right)\right)}^{2^{\beta }}.$$

If $\sigma \left(x^{2r}\right),$$\sigma \left({\left(x+1\right)}^{2r\prime }\right),$$\sigma \left(x^{u-1}\right)$, and $\sigma \left({\left(x+1\right)}^{v-1}\right)$ are Mersenne, then $2r,2r^{\prime },u-1,v-1\in \{2,4\}.$ Thus, $a=b=4.$ If $\sigma \left(x^{2r}\right),$$\sigma \left(x^{u-1}\right),$$\sigma \left({\left(x+1\right)}^{2r\prime }\right)$ and $\sigma \left({\left(x+1\right)}^{v-1}\right)$ are not Mersenne, then $r,r^{\prime },u-1,v-1>2$ and $\omega \left({\sigma }^{**2}\left(A\right)\right)>2,$ a contradiction. For $a=b=2,$ A is bi-unitary perfect; hence, A is a bi-unitary superperfect.

(4)

Now, let $a=2m+1$ and $b=2n$. Since ${\sigma }^{**}\left({\left(x+1\right)}^{2n}\right)$ splits over ${\mathbb{F}}_2$ only when $n=1$, then ${\sigma }^{**2}\left(A\right)={\sigma }^{**}\left({\sigma }^{**}\left(x^{2m+1}\right){\sigma }^{**}\left({(x+1)}^2\right)\right).$ However, ${\sigma }^{**}\left(x^{2m+1}\right)$ splits over ${\mathbb{F}}_2$ if $2m+1$ is of the form $2^d-1.$ If $a=2m$ and $b=2n+1,$ then $a=2$ and $b=2^d-1.$ The sufficient condition can be easily verified.

□

The proof of Theorem 2 is now complete.

7. Some Values of ${\mathit{\sigma }}^{\mathbf{*}\mathbf{*}}\mathbf{\left(}\mathit{A}\mathbf{\right)}$ and ${\mathit{\sigma }}^{\mathbf{*}\mathbf{*}\mathbf{2}}\mathbf{\left(}\mathit{A}\mathbf{\right)}$

For convenience of readers, we list the below table (Table 1) that consists of the values of ${\sigma }^{**}\left(A\right)$ and ${\sigma }^{**2}\left(A\right)$ for $A\in \{x^a,{(x+1)}^a,M_i^b\}$, where $1\le \mathrm{a}\le 13$, $1\le b\le 7$. We consider the polynomials $C_1=x^4+x+1$, $C_2=x^6+x^5+x^4+x^2+1$, $C_3=x^6+x^5+x^4+x+1$, and $C_4=x^{10}+x^9+x^8+x^7+x^2+x+1$.

Table 1. $A\in \{x^a,{(x+1)}^a,M^a\}$.

A	a	${\mathit{\sigma }}^{**}$	${\mathit{\sigma }}^{**2}$
$x^a$	1	x	$x+1$
	2	$x^2$	${(x+1)}^2$
	3	$x^3$	${(x+1)}^3$
	4	$x^2{M}_1$	$x{(x+1)}^3$
	5	$x{M}_1^2$	$x^2{(x+1)}^3$
	6	$x^4{M}_1$	$x{(x+1)}^3{M}_1$
	7	$x^7$	${(x+1)}^7$
	8	$x^4{M}_5$	$x^3{(x+1)}^3{M}_1$
	9	$x{M}_5^2$	$x^6{(x+1)}^3$
	10	$x^2{M}_1^2{M}_5$	$x^5{(x+1)}^5$
	11	$x^3{M}_1^4$	$x^2{(x+1)}^5{C}_1$
	12	$x^2{M}_1^2{M}_2{M}_3$	$x^5{(x+1)}^7$
	13	$x{M}_2^2{M}_3^2$	$x^6{(x+1)}^7$
................	................	................	................
${(1+x)}^a$	1	x	$x+1$
	2	$x^2$	${(x+1)}^2$
	3	$x^3$	${(x+1)}^3$
	4	$x^2{M}_1$	$x{(x+1)}^3$
	5	$x{M}_1^2$	$x^2{(x+1)}^3$
	6	$x^4{M}_1$	$x{(x+1)}^3{M}_1$
	7	$x^7$	${(x+1)}^7$
	8	$x^4{M}_5$	$x^3{(x+1)}^3{M}_1$
	9	$x{M}_5^2$	$x^6{(x+1)}^3$
	10	$x^2{M}_1^2{M}_5$	$x^5{(x+1)}^5$
	11	$x^3{M}_1^4$	$x^2{(x+1)}^5{C}_1$
	12	$x^2{M}_1^2{M}_2{M}_3$	$x^5{(x+1)}^7$
	13	$x{M}_2^2{M}_3^2$	$x^6{(x+1)}^7$
................	................	................	................
$M_1^a$	1	$x(x+1)$	$x(x+1)$
	2	$x^2{(x+1)}^2$	$x^2{(x+1)}^2$
	3	$x^3{(x+1)}^3$	$x^3{(x+1)}^3$
	4	$x^2{(x+1)}^2{C}_1$	$x^3{(x+1)}^3{M}_1$
	5	$x(x+1)C_1^2$	$x^3{(x+1)}^3{M}_1^2$
	6	$x^4{(x+1)}^4{C}_1$	$x^3{(x+1)}^3{M}_1^3$
	7	$x^7{(x+1)}^7$	$x^7{(x+1)}^7$
................	................	................	................
$M_2^a$	1	$x{(x+1)}^2$	$x^2(x+1)$
	2	$x^2{(x+1)}^4$	$x^2{(x+1)}^2{M}_1$
	3	$x^3{(x+1)}^6$	$x^4{(x+1)}^3{M}_1$
	4	$x^2{(x+1)}^4{M}_1{M}_5$	$x^6{(x+1)}^4{M}_1$
	5	$x{(x+1)}^2{M}_1^2{M}_5^2$	$x^{10}{(x+1)}^5$
	6	$x^4{(x+1)}^8{M}_1{M}_5$	$x^8{(x+1)}^4{M}_1{M}_5$
	7	$x^7{(x+1)}^{14}$	$x^8{(x+1)}^7{M}_2{M}_3$
$M_3^a$	1	$x^2(x+1)$	$x{(x+1)}^2$
	2	$x^4{(x+1)}^2$	$x^2{(x+1)}^2{M}_1$
	3	$x^6{(x+1)}^3$	$x^3{(x+1)}^4{M}_1$
	4	$x^4{(x+1)}^2{M}_1{M}_4$	$x^4{(x+1)}^6{M}_1$
	5	$x^2(x+1)M_1^2{M}_4^2$	$x^5{(x+1)}^{10}$
	6	$x^8{(x+1)}^4{M}_1{M}_4$	$x^4{(x+1)}^8{M}_1{M}_4$
	7	$x^{14}{(x+1)}^7$	$x^7{(x+1)}^8{M}_2{M}_3$
................	................	................	................
$M_4^a$	1	$x{(x+1)}^3$	$x^3(x+1)$
	2	$x^2{(x+1)}^6$	$x^4{(x+1)}^2{M}_1$
	3	$x^3{(x+1)}^9$	$x{(x+1)}^3{\left(M_5\right)}^2$
	4	$x^2{(x+1)}^6{M}_1{C}_2$	$x^7{(x+1)}^4{M}_1{M}_2$
	5	$x{(x+1)}^3{M}_1^2{C}_2^2$	$x^9{(x+1)}^5{M}_2^2$
	6	$x^4{(x+1)}^{12}{M}_1{C}_2$	$x^5{(x+1)}^4{M}_1^3{M}_2^2{M}_3$
	7	$x^7{(x+1)}^{21}$	$x{(x+1)}^7$
			$C_4^2$
................	................	................	................
${\left(M_5\right)}^a$	1	$x^3(x+1)$	$x{(x+1)}^3$
	2	$x^6{(x+1)}^2$	$x^2{(x+1)}^4{M}_1$
	3	$x^9{(x+1)}^3$	$x^3(x+1)M_4^2$
	4	$x^6{(x+1)}^2{M}_1{C}_3$	$x^4{(x+1)}^7{M}_1{M}_3$
	5	$x^3(x+1)M_1^2{C}_3^2$	$x^5{(x+1)}^9{M}_3^2$
	6	$x^{12}{(x+1)}^4{M}_1{C}_3$	$x^4{(x+1)}^5{M}_1^3{M}_2{M}_3^2$
	7	$x^{21}{(x+1)}^7$	$x^7(x+1){\left(\sigma \left(x^{10}\right)\right)}^2$

8. Conclusions

In conclusion, we proved the non-existence of odd bi-unitary superperfect polynomials and provided a classification for bi-unitary superperfect polynomials over ${\mathbb{F}}_2$ based on their irreducible factors. In particular, we showed that a non-constant bi-unitary superperfect polynomial A over ${\mathbb{F}}_2$ can be divisible by one irreducible polynomial x or $x+1$ with exponent 2 or $2^n-1$ for a positive integer n. Furthermore, we showed that the only bi-unitary superperfect polynomials over ${\mathbb{F}}_2$ with exactly two irreducible factors are of the form $x^a{(x+1)}^b$ with $a,b\in \{2,4,9,13,2^d-1\}$, d is a positive integer.