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

1. Introduction

Let n and k be positive integers and let $\sigma \left(n\right)$(resp. ${\sigma }^{*}\left(n\right)$) denotes 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 ${\sigma }^k\left(n\right)=\underset{k-\mathrm{times}}{\underbrace{\sigma \left(\sigma \right(...\left(\sigma \right(}}n\left)\right)\left)\right)=2n$. 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, 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 $n\in \{2,9\}$.

Now, let A be a nonzero polynomial defined over the prime field ${\mathbb{F}}_2$. 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.

A is an even polynomial if it has a linear factor in ${\mathbb{F}}_2\left[x\right]$ else 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,$$M_4=1+x+x^2+x^3+x^4,$$M_5=1+x^3+x^4.$ Note that all these polynomials are irreducible, so we call them Mersenne primes.

Let $\omega \left(A\right)$ denotes the number of distinct irreducible monic polynomials that divide A. The notion of perfect polynomials 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. ${\sigma }^{**}\left(A\right)=A$). Also, 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. So, working on bi-unitary superperfect polynomial over ${\mathbb{F}}_2$ is not an easy task especially when A is divisible by more than 2 irreducible factors.

Many researchers studied the unitary perfect polynomials over ${\mathbb{F}}_2$. The authors in [7,9,10] list the unitary perfect polynomials over ${\mathbb{F}}_2$ with $\omega \left(A\right)\le 4.$ They list others that are divisible by $x(x+1)P,$P is a Mersenne polynomial, raised to certain powers, see [7]). Beard [11] found many bi-unitary perfect polynomials over $F_{p^d}$, some of which are neither perfect nor unitary perfect. He conjectured a characterization of the bi-unitary perfect polynomials which splits over $F_p$ when $p>2.$ Beard gave examples of non-splitting bi-unitary perfect polynomials over $F_p$ when $p\in \{2,3,5\}$. Rahavandrainy [12] gave all bi-unitary perfect polynomials over the prime field ${\mathbb{F}}_2$, with at most four irreducible factors (Lemmas 12 and 13). Gallardo and Rahavandrainy [13] classified some unitary superperfect polynomials with a small number of prime divisors under some conditions on the number of prime factors of ${\sigma }^{*}\left(A\right)$.

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.

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, (Theorem 1). 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 $x^2{(x+1)}^{2^n-1}$ are such examples, n is a positive integer.

Our main result is given in the following theorem:

Theorem 1.

If $\omega \left(A\right)\le 2$ and A is a bi-unitary superperfect over ${\mathbb{F}}_2$ if and only if $A,\overline{A}\in \{x^2,x^{2^d-1},x^2{(x+1)}^2,x^4{(x+1)}^4,x^9{(x+1)}^9,x^9{(x+1)}^{13},x^2{(x+1)}^{2^d-1},x^{2^m-1}{(x+1)}^{2^n-1}\},$ where $m,n\in {\mathbb{N}}^{*}.$

2. Preliminaries

The following two lemmas are helpful.

Lemma 1.

Let A be a polynomial in ${\mathbb{F}}_2\left[x\right]$, then ${\sigma }^{*}\left(A^{2^n}\right)={\left({\sigma }^{*}\left(A\right)\right)}^{2^n},$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 [13]] 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 [11], and Rahavandrainy [12] works.

Lemma 4.

[Theorem 1 and its Corollary in [11]] If A is a non-constant bi-unitary perfect polynomial, then $x(x+1)$ divides A and $\omega \left(A\right)$$\ge 2.$

Lemma 5.

[Lemma 2.2 in [12]]

1)

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

2)

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

Corollary 1.

[Corollary 2.3 in [12]] 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 lemma follows from Lemma 5 and the binomial formula. Table Section 6 shows some values of ${\sigma }^{**}\left(A\right)$ when A is a power of the first five Merssene primes.

Lemma 6.

Let the polynomial $M_i$ be 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)

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

3)

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

4)

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

5)

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

6)

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

Lemma 7.

[Corollary 2.4 in [12]]

1)

${\sigma }^{**}\left(x^a\right)$ splits over $F_2$ 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 $F_2$ if and only if P is Mersenne and $c=2$ or $c=2^d-1$ for some $d\in {\mathbb{N}}^{*}$.

Lemma 8 summarizes key results taken from Canaday’s paper [5].

Lemma 8.

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 9.

[Lemma 2.6 in [14]] 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 squarefree.

3. 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 proof of the following lemmas follow directly.

Lemma 10.

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

Lemma 11.

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 [12]) 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.

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

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

Lemma 12.

[Theorem 1.1 in [12]] Let $A\in {\mathbb{F}}_2\left[x\right]$ be bi-unitary perfect polynomial such that $\omega \left(A\right)=3$. Then $A,\overline{A}\in \{B_j:j\le 7\}.$

Lemma 13.

[Theorem 1.2 in [12]] Let $A\in {\mathbb{F}}_2\left[x\right]$ be bi-unitary perfect polynomial such that $\omega \left(A\right)=4$. Then $A,\overline{A}\in \{B_j:8\le j\le 13\}\bigcup \{R_1,R_2\}.$

Lemma 14.

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

4. Proof of Theorem 1

We start this section by the following corollary.

Corollary 2.

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 Lemma 5. □

Lemma 15.

$x(x+1)$ 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 ${\sigma }^{**}\left(P^{2n+1}\right)\left(0\right)=1+\underset{(2n+1)-\mathrm{times}}{\underbrace{P\left(0\right)+\dots +P^{2n+1}\left(0\right)}}=1+2n+1=0$. If $a=2n$, then $1+P^{n+1}\left(0\right)=0.$ So, 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 16.

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 $gcd(P,B)=1.$ But, $1+P$ 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 we are done by Corollary 2. 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 by Lemma 15. □

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 16). So, $x(x+1)$ divides ${\sigma }^{**}\left(B\right)={\sigma }^{**2}\left(A\right)$. □

The following lemma is similar to Lemma 7.

Lemma 17.

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, the $a=2$ (Lemma 7) and ${\sigma }^{**2}\left(x^a\right)={(x+1)}^2$. Suppose, ${\sigma }^{**}\left(x^a\right)$ 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, $r\ge 1$. But ${\sigma }^{**2}\left(x^a\right)={\sigma }^{**}\left({(1+x)}^{2^{\alpha }}·\sigma \left(x^{2r}\right)·{\left(\sigma \left(x^{u-1}\right)\right)}^{2^{\alpha }}\right)$, so ${\sigma }^{**}\left({(1+x)}^{2^{\alpha }}\right)$ must split. Hence, $\alpha =1$ and since $\sigma \left(x^{2r}\right)$ is odd and square free (Lemma 9), then $\sigma \left(x^{2r}\right)$ has a Mersenne factor. So, $2r\le 6$ 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 (Lemma 7). 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, by Lemma 9, $\sigma \left(x^{2r}\right)$ has a Mersenne factor. So, $u-1\le 6$ 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 done for ${\sigma }^{**2}\left({(x+1)}^a\right)$ and the proof is compete. □

Lemma 18.

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

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

□

4.1. Case w(A)=1

We prove that ${\sigma }^{**}\left(A\right)$ can not have more than one prime factor when A is a prime power.

Lemma 19.

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 17. □

Lemma 20.

If $A=P^{\alpha }\in {\mathbb{F}}_2\left[x\right]$, then A 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 by Lemma 16 we have $x(x+1)$ 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 5.

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.

Theorem 2.

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. 

By Corollary 5, $A=x^{\alpha }$ 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 by a direct computation and the result follows since if A is a bi-unitary superperfect, then so is $\overline{A}$. □

4.2. Case w(A)=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 ${\mathbb{F}}_2$, see Lemma 10 and (Theorem 5 in [11]$).$

Corollary 6.

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

Proof. 

Follows directly from Corollary 4. □

Lemma 21.

[Lemma 3.1 in [12]] 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 22.

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 ${\sigma }^{**}\left(x^a{(x+1)}^b\right)$ doe 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 is odd and b is even, 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 ${\sigma }^{**2}\left(A\right)={\sigma }^{**}\left({\sigma }^{**}\left(x^a\right){(1+x)}^b\right).$ But ${\sigma }^{**}\left(x^{2m+1}\right)$ and ${\sigma }^{**}{(x+1)}^{2n+1}$ split over ${\mathbb{F}}_2$ when $2m+1$ and $2n+1$ are of the form $2^d-1$ (Lemma 7).

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. ${\sigma }^{**}\left(x^a{(x+1)}^b\right)={\sigma }^{**}\left({(1+x)}^{2^{\alpha }-1}{\left(\sigma \left(x^{u-1}\right)\right)}^{2^{\alpha }}{x}^{2^{\beta }-1}\sigma {\left({\left(x+1\right)}^{v-1}\right)}^{2^{\beta }}\right).$ By Lemma 21$(u-1\ge 3$ and $\alpha =1)$ or $(v-1\ge 3$ and $\beta =1)$. Also, $\sigma \left(x^{u-1}\right)$ 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. So, there exit 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 $u,v\in \left\{5,7\right\}$. 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 $2^{\alpha }u-1$ with u is odd and $b\in \{4r^{\prime },4r^{\prime }+2\}$ such that $2r^{\prime }-1$, $2r^{\prime }+1$ is of the form $2^{\beta }v-1$, 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 $2r,2r^{\prime },u-1,v-1\in \{2,4\}.$ So, $a=b=4.$ 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 and 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 ${\sigma }^{**2}\left(A\right)={\sigma }^{**}\left({\sigma }^{**}\left(x^{2m+1}\right){\sigma }^{**}\left({(x+1)}^2\right)\right).$ But ${\sigma }^{**}\left(x^{2m+1}\right)$ 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 1 is now complete.

5. Conclusion

In conclusion, a non constant bi-unitary superperfect polynomial A over ${\mathbb{F}}_2$ can be divisible by one irreducible polynomial x or $x+1$ and its exponent is 2 or $2^n-1$ for a positive integer n. Moreover, the only bi-unitary superperfect polynomials over ${\mathbb{F}}_2$ with exactly two prime factors are $x^a{(x+1)}^b$ with $a,b\in \{2,4,9,13,2^d-1\}$, d is a positive integer and $a=b$ if and only if $a,b\in \{2,4\}$.

6. Table

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 $C_4=x^{10}+x^9+x^8+x^7+x^2+x+1$. The below table lists the values of ${\sigma }^{**}\left(A\right)$ and ${\sigma }^{**2}\left(A\right)$ for $A\in \{x^a,{(x+1)}^a,M_i^b\}$ with $1\le \mathrm{a}\le 13$, $1\le b\le 7$.