On k−Unitary Perfect Polynomials over F2

1. Introduction

The study of perfect numbers and their analogues-unitary perfect and bi-unitary perfect numbers has a long and rich history in classical number theory. A classical result of Euclid [12] shows that if $2^p-1$ is prime (a Mersenne prime), then $2^{p-1}(2^p-1)$ is an even perfect number. Despite centuries of research, the existence of odd perfect numbers remains an open problem. Similarly, only a few unitary perfect numbers are known (such as $6,60,90,87360,\dots$), and the existence of infinitely many remains open. Importantly, no odd unitary perfect integer has been discovered.

Motivated by these classical problems, several authors [1,2,3,4,8,9] have investigated polynomial analogues of perfectness over finite fields. In particular, perfect, unitary perfect, and bi-unitary perfect polynomials over ${\mathbb{F}}_2$ have been studied extensively, leading to strong structural restrictions and complete classifications in certain cases.

Let $A\in {\mathbb{F}}_2\left[x\right]$ be a nonzero polynomial. 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 $gc{d}_u$ denotes the greatest common unitary divisor. The sum of divisors (resp. unitary divisors and bi-unitary divisors) of A is denoted by $\sigma \left(A\right)($resp. ${\sigma }^{*}\left(A\right)$ and ${\sigma }^{**}\left(A\right))$. In symbols,

$$\sigma \left(A\right)=\sum _{B|A}B,{\sigma }^{*}\left(A\right)=\underset{gcd(B,A/B)=1}{\sum _{B|A}}B,\mathrm{and}{\sigma }^{**}\left(A\right)=\underset{gc{d}_u(B,A/B)=1}{\sum _{B|A}}B.$$

A polynomial $A\in {\mathbb{F}}_2\left[x\right]$ is perfect (resp. unitary perfect and bi-unitary perfect) if $\sigma \left(A\right)=A$ (resp. ${\sigma }^{*}\left(A\right)=A$ and ${\sigma }^{**}\left(A\right)=A$). Note that in studying perfect or unitary perfect polynomials, the distinction disappears when the polynomial is squarefree.

Also, A is called a bi-unitary superperfect if ${\sigma }^{**2}\left(A\right)={\sigma }^{**}\left({\sigma }^{**}\left(A\right)\right)=A$. Chehade et al. [7] gave all bi-unitary superperfect polynomials divisible by one or two irreducible polynomials over ${\mathbb{F}}_2$. Note that the functions $\sigma$, ${\sigma }^{*}$ and ${\sigma }^{**}$ are degree preserving and multiplicative while ${\sigma }^{**2}$ is a degree preserving function but not a multiplicative one.

2. Preliminary

Throughout this paper, all polynomials are assumed to be monic in ${\mathbb{F}}_2\left[x\right]$. For a polynomial A, we denote by $deg\left(A\right)$ its degree and by $\omega \left(A\right)$ the number of its distinct irreducible factors. Unless otherwise stated, the following notations are used.

• $\mathbb{N}$(resp. ${\mathbb{N}}^{*})$ represents the set of non-negative (resp. positive) integers.
• $\overline{A}$ is the polynomial obtained from A with x replaced by $x+1$, that is $\overline{A}\left(x\right)=A(x+1)$.
• $A^{*}$ is the inverse of the polynomial A with $deg\left(A\right)=m$, in this sense $A^{*}\left(x\right)=x^{m}A\left({\displaystyle \frac{1}{x}}\right)$.

A polynomial of the form $T=1+x^a{(x+1)}^b$ with $gcd(a,b)=1$ is called a Mersenne polynomial. If such a polynomial is irreducible over ${\mathbb{F}}_2$, it is called a Mersenne prime. The first Mersenne primes over ${\mathbb{F}}_2$ are

$$T_1=1+x+x^2,T_2=1+x+x^3,T_3=1+x^2+x^3,T_4=1+x+x^2+x^3+x^4,T_5=1+x^3+x^4.$$

A polynomial in ${\mathbb{F}}_2\left[x\right]$ is called even if it is divisible by x or $x+1$; otherwise, it is called odd.

Recently, the divisor function $\sigma$was generalized by summing $k-$th powers of divisors, giving rise to the notion of $k-$perfect polynomials. Chehade et al. [6] introduced a polynomial analogue of ${\sigma }_k\left(n\right),$$n\in \mathbb{N}$, by defining for any nonzero polynomial $A\in {\mathbb{F}}_2\left[x\right]$ as follows: ${\sigma }_k\left(A\right)=\sum _{B\mid A}{B}^k.$ If $A=\prod _{i=1}^m{P}_i^{{\alpha }_i},$ then

$${\sigma }_k\left(A\right)=\prod _{i=1}^m\left({\displaystyle \frac{P_i^{k({\alpha }_i+1)}-1}{P_i^k-1}}\right).$$

Chehade et al. [6] also defined $k-$perfect polynomials over ${\mathbb{F}}_2$ as those polynomials A satisfying ${\sigma }_k\left(A\right)=A^k$. They showed that no odd $2^n-$perfect polynomials exist over ${\mathbb{F}}_2$ and characterized all even $2^n-$perfect polynomials over ${\mathbb{F}}_2$ that have the form $x^a{(x+1)}^b\prod _{i=1}^m{P}_i^{h_i}$, where each $P_i$ is Mersenne and $a,b,$ and $h_i$ are positive integers.

2.1. $k-$Unitary Perfect Polynomials

Let A be a nonzero polynomial in ${\mathbb{F}}_2\left[x\right].$ Define ${\sigma }_k^{*}\left(A\right)$ to be the sum of the $k-$th powers of its distinct unitary divisors. That is, ${\sigma }_k^{*}\left(A\right)=\underset{gcd(B,A/B)=1}{\sum _{B|A}}{B}^k$ and if $A=\prod _{i=1}^m{P}_i^{h_i},$ then

$${\sigma }_k^{*}\left(A\right)=\prod _{i=1}^m\left(P_i^{k{\alpha }_i}+1\right).$$

The function ${\sigma }_k^{*}$ is multiplicative and degree preserving.

We now introduce the central concept investigated in this work, namely $k-$unitary perfect polynomials over ${\mathbb{F}}_2$.

Definition 1. 

Let k be a positive integer. A polynomial A is called $k-$unitary perfect over ${\mathbb{F}}_2$ if

$${\sigma }_k^{*}\left(A\right)=A^k.$$

Example 1. 

Let $A\left(x\right)=x^2{(x+1)}^3(x^2+x+1)\in {\mathbb{F}}_2\left[x\right]$, then

$$\begin{array}{cc}\hfill {\sigma }_4^{*}\left(A\left(x\right)\right)& =(1+x^{2\left(4\right)})(1+{(x+1)}^{3\left(4\right)})(1+{(x^2+x+1)}^4)\hfill \\ \hfill & =x^8{(x+1)}^{12}{(x^2+x+1)}^4\hfill \\ \hfill & =A^4.\hfill \end{array}$$

Hence, A is a $4-$unitary perfect polynomial over ${\mathbb{F}}_2.$

The unitary perfect polynomial is studied extensively by Beard and Habrin [1,2,3] and by Gallardo and Rahavandrainy [9] and [10].

When $k=1$, Definition 1 reduces to the classical notion of unitary perfect polynomials. In this work, we focus primarily on the case $k=2^n$$(n\in {\mathbb{N}}^{*}),$ where the arithmetic of ${\mathbb{F}}_2$ plays a crucial role. We classify all $2^n-$unitary perfect polynomials over ${\mathbb{F}}_2$ with at most three distinct irreducible factors, proving that no odd such polynomials exist and fully characterizing the even ones of the form $x^a{(x+1)}^b{P}^h,$ where P is a Mersenne prime and $a,b$ and h are positive integers.

Our main result is given in the following theorem.

Theorem 1. 

Let $a,b,h\in \mathbb{N}$ and let P be a Mersenne prime in ${\mathbb{F}}_2\left[x\right]$. Then, $A=x^a{(x+1)}^b{P}^h$ is a $2^n-$unitary perfect polynomial over ${\mathbb{F}}_2$ if and only if

• $A=x^{2^t}{(x+1)}^{2^t}$ if $\omega \left(A\right)=2,$
• $A\in \left\{x^{3.2^t}{(x+1)}^{2^{t+1}}{T}_1^{2^t},x^{2^{t+1}}{(x+1)}^{3.2^t}{T}_1^{2^t},x^{3.2^t}{(x+1)}^{3.2^t}{T}_1^{2^{t+1}},x^{5.2^t}{(x+1)}^{2^{t+2}}{T}_4^{2^t},\right.$
  $\left.x^{2^{t+2}}{(x+1)}^{5.2^t}{T}_5^{2^t}\right\}$ if $\omega \left(A\right)=3,$

for some positive integer $t.$

The family $\{x^{2^t}{(x+1)}^{2^t}:t\in \mathbb{N}\}$ may be regarded as an analogue of the family $\{x^{2^t-1}{(x+1)}^{2^t-1}:t\in \mathbb{N}\}$, which corresponds to the trivial instance of perfect polynomials over ${\mathbb{F}}_2$.

3. Useful Results

We consider the following families of polynomials arising in the computation of ${\sigma }^{*}\left(x^{p^t}\right),$$p\in \left\{3,5,7,11\right\}$:

$$\begin{array}{c}{S}_t=1+x^{3^t}+x^{2.3^t},\hfill \\ R_t=1+x^{7^t}+x^{3.7^t},\hfill \\ B_t=1+x^{2.7^t}+x^{3.7^t},\hfill \\ C_t=1+x^{5^t}+x^{2.5^t}+x^{3.5^t}+x^{4.5^t},\hfill \\ D_t=1+x^{{11}^t}+x^{2.{11}^t}+...+x^{10.{11}^t}.\hfill \end{array}$$

Note that the polynomials $S_t,$$B_t,$$C_t,$$R_t$ and $D_t$ are irreducible over ${\mathbb{F}}_2$ with $S_0=T_1,R_0=T_2,B_0=T_3,C_0=T_4.$

We need the following useful results. Since some of them are straightforward, their proofs are omitted.

Lemma 1. 

[9] If $A=A_1{A}_2$ is unitary perfect over ${\mathbb{F}}_2$ and if $gcd(A_1,A_2)=1$. Then $A_1$ is unitary perfect if and only if $A_2$ is unitary perfect.

Lemma 2. 

[9] If $A\left(x\right)$ is unitary perfect over ${\mathbb{F}}_2$, then the polynomial $A(x+1)$ is also unitary perfect over ${\mathbb{F}}_2$.

Lemma 3. 

[9] If $A\left(x\right)$ is unitary perfect over ${\mathbb{F}}_2$, then $A^{2^n}$ is also unitary perfect over ${\mathbb{F}}_2$, for any $n\in \mathbb{N}$.

In [9], Gallardo and Rahavandrainy gave a complete list for all even perfect polynomials with 3 irreducible factors as given in the following lemma.

Lemma 4. 

The complete list of even perfect polynomials over ${\mathbb{F}}_2$ with $\omega \left(A\right)=3$ is of the form $A^{2^n}$or ${\overline{A}}^{2^n}$where $A\in \left\{x^2{(x+1)}^3{T}_1,x^3{(x+1)}^3{T}_1^2,x^4{(x+1)}^5{T}_4\right\}$.

The statement in part iii) of the following lemma is a result of Dickson (see [5])

Lemma 5. 

i) Any complete polynomial inverts into itself.

ii) If $1+x+\dots +x^h=PQ$, then either $(P=P^{*},Q=Q^{*})$ or $(P=Q^{*},Q=P^{*})$.

iii) If $P=P^{*}$ is Mersenne, then $P\in \{T_1,T_4\}.$

Corollary 1. 

If $P=\overline{P}$ is Mersenne, then $P=T_1.$

Lemma 6.([5], Lemmata 4, 5, 6 and Theorem 8) Let $P,Q\in {\mathbb{F}}_2\left[x\right]$ and let $n,m\in \mathbb{N}$.

i) If $1+P+\dots +P^{2n}=Q^m$, then $m\in \{0,1\}$.

ii) If $1+P+\dots +P^{2n}=Q^{m}A$, with $m>1$ and $A\in {\mathbb{F}}_2\left[x\right]$ is non-constant, then $\mathrm{deg}\left(P\right)>\mathrm{deg}\left(Q\right)$.

iii) If $1+x+\dots +x^{2n}=PQ$ and $P=1+(x+1)+\dots +{(x+1)}^{2m}$, then $n=4$, $P=T_1$ and $Q=P\left(x^3\right)=1+x^3+x^6$.

iv) If any irreducible factor of $1+x+\dots +x^{2n}$ is Mersenne, then $n\in \{1,2,3\}$.

v) If $1+x+\dots +x^h=1+(x+1)+\dots +{(x+1)}^h$, then $h=2^n-2$, for some $n\in \mathbb{N}$.

Lemma 7. 

i) 

If $T_1$ divides $1+x+\dots +x^h$, then $h\equiv 2$$(mod$$3)$.

ii) 

If $T_2$ or $T_3$ divides $1+x+\dots +x^h$, then $h\equiv 6$ $(mod$$7)$.

iii) 

If $T_4$ or $T_5$ divides $1+x+\dots +x^h$, then $h\equiv 4$$(mod$$5)$.

As a special case of ([11], Theorem 2.47), we have

Lemma 8. 

The polynomial $1+x+\dots +x^m$ is irreducible over ${\mathbb{F}}_2$ if and only if $m+1$ is a prime number and 2 is a primitive root in ${\mathbb{F}}_{m+1}$.

Consequently, we get

Lemma 9. 

i) The polynomial $Q\left(x\right)=1+x^5+\dots +{\left(x^5\right)}^l$ is irreducible over ${\mathbb{F}}_2$ if and only if $l=4$.

ii) 

The polynomial $Q\left(x\right)=1+x+\dots +x^{3.2^r}$ is irreducible over ${\mathbb{F}}_2$ if and only if $r=2$.

iii) 

The polynomial $Q\left(x\right)=1+x+\dots +x^{5.2^r}$ is irreducible over ${\mathbb{F}}_2$ if and only if $r=1$.

Lemma 10. 

[9] Any non-constant unitary perfect polynomial over ${\mathbb{F}}_2$ is divisible by x and by $x+1$. In particular, there is no odd unitary perfect polynomial over ${\mathbb{F}}_2$.

Lemma 11. 

If P is a Mersenne prime, then for any $n\in {\mathbb{N}}^{*}$, ${\sigma }^{*}\left(P^n\right)$ has always a factor $x(x+1).$

Lemma 12. 

Let P be irreducible and let $A=P^{\alpha }\in {\mathbb{F}}_2\left[x\right]$ with $\alpha \ge 1.$ Then ${\sigma }_k^{*}\left(A\right)={\left({\sigma }^{*}\left(A\right)\right)}^k$ if and only if $k=2^n.$

Proof. 

${\sigma }_k^{*}\left(A\right)={\left({\sigma }^{*}\left(A\right)\right)}^k$ gives

$$1+P^{k\alpha }={\sigma }_k^{*}\left(A\right)={\left(1+P^{\alpha }\right)}^k.$$

In ${\mathbb{F}}_2\left[x\right],$ the binomial expansion of ${\left(1+P^{\alpha }\right)}^k$ reduces to $1+P^{k\alpha }$ if and only if all intermediate binomial coefficients vanish modulo 2, which occurs precisely when k is a power of 2. □

Corollary 2. 

Let $A=\prod _{i=1}^r{P}_i^{{\alpha }_i}\in {\mathbb{F}}_2\left[x\right]$ where $P_i$ is irreducible, then ${\sigma }_{2^n}^{*}\left(A\right)={\left({\sigma }^{*}\left(A\right)\right)}^{2^n}.$

The following corollary is a direct consequence of Corollary .

Corollary 3. 

$A=\prod _{i=1}^m{P}_i^{{\alpha }_i}$ is unitary perfect over ${\mathbb{F}}_2$ iff A is $2^n-$unitary perfect over ${\mathbb{F}}_2.$

Lemma 13. 

Let $m\le n$ be positive integers and let $A\in {\mathbb{F}}_2\left[x\right],$ then ${\sigma }_{2^m}^{*}\left(A\right)$ divides ${\sigma }_{2^n}^{*}\left(A\right).$

Proof. 

$$\begin{array}{cc}\hfill {\sigma }_{2^n}^{*}\left(A\right)& ={\left({\sigma }^{*}\left(A\right)\right)}^{2^n}\hfill \\ \hfill & ={\left({\sigma }^{*}\left(A\right)\right)}^{2^m}{\left({\sigma }^{*}\left(A\right)\right)}^{2^{n-m}}\hfill \\ \hfill & ={\sigma }_{2^m}^{*}\left(A\right){\left({\sigma }^{*}\left(A\right)\right)}^{2^{n-m}}.\hfill \end{array}$$

□

Hence if A is a multi-perfect polynomial over ${\mathbb{F}}_2,$$(A$ divides $\sigma \left(A\right))$, then $A$is a $2^n-$multi-perfect polynomial over ${\mathbb{F}}_2$.

Lemma 14. 

Assume $\alpha \ge 1,$P is an irreducible polynomial and $A=P^{\alpha }\in {\mathbb{F}}_2\left[x\right],$ then $A\nmid {\sigma }_k^{*}\left(A\right).$

Proof. 

Assume that $A|{\sigma }_k^{*}\left(A\right)$, then there exists $Q\in {\mathbb{F}}_2\left[x\right]$ such that ${\sigma }_k^{*}\left(A\right)=QA$ with $deg\left(Q\right)<deg\left(A^k\right)$. So, $1+P^{k\alpha }=Q{P}^{\alpha }$ and $P^{\alpha }(Q+P^{k\alpha -\alpha })=1.$ This means that $P^{\alpha }$ is a unit in ${\mathbb{F}}_2\left[x\right].$ Hence, $P^{\alpha }=1$. This contradicts the fact that P is irreducible in ${\mathbb{F}}_2\left[x\right]$. □

Corollary 4. 

Let $\alpha \ge 1$ and let $A=P^{\alpha }\in {\mathbb{F}}_2\left[x\right]$, then A is not $k-$unitary perfect, for any irreducible P.

The proof of Corollary 4 can be directly obtained from Lemma 11 or from Corollary .

As a result of corollary 14 and lemma , we have $A\nmid {\sigma }_k^{*}\left(A\right)$ and hence ${\sigma }_k^{*}\left(A\right)\nmid A^k.$

Lemma 15. 

Let $A_1,\dots ,A_m\in {\mathbb{F}}_2\left[x\right]$. Then

$$\prod _{i=1}^m{A}_i(x+1)=\left(\prod _{i=1}^m{A}_i\right)(x+1).$$

Proof. 

Define the map $\phi :{\mathbb{F}}_2\left[x\right]\to {\mathbb{F}}_2\left[x\right]$ by

$$\phi \left(f\right(x\left)\right)=f(x+1).$$

Since substitution preserves addition and multiplication, $\phi$ is a ring automorphism of ${\mathbb{F}}_2\left[x\right]$. Hence,

$$\phi \left(\prod _{i=1}^m{A}_i\left(x\right)\right)=\prod _{i=1}^m\phi \left(A_i\left(x\right)\right)=\prod _{i=1}^m{A}_i(x+1).$$

By definition of $\phi$, the left-hand side equals

$$\left(\prod _{i=1}^m{A}_i\right)(x+1),$$

which proves the result. □

Lemma 16. 

If $A\left(x\right)$ is $2^n-$unitary perfect over ${\mathbb{F}}_2$, then the polynomial $\overline{A}\left(x\right)$ is also $2^n-$unitary perfect over ${\mathbb{F}}_2$.

Proof. 

Let $A\left(x\right)={\prod }_{i=1}^m{P}_i{\left(x\right)}^{a_i},$where the $P_i\left(x\right)\in {\mathbb{F}}_2\left[x\right]$ are distinct irreducible polynomials. Since $A\left(x\right)$ is $2^n-$unitary perfect, we have

$${\sigma }_{2^n}^{*}\left(A\right)=\prod _{i=1}^m\left(1+{P_i}^{2^n{a}_i}\right)=A^{2^n}.$$

Let $B\left(x\right)=A(x+1)={\prod }_{i=1}^m{P}_i{(x+1)}^{a_i}.$ Since the substitution $x\mapsto x+1$ is a ring automorphism of ${\mathbb{F}}_2\left[x\right]$, it preserves irreducibility and coprimality. Hence, the polynomials $P_i(x+1)$ are distinct irreducible polynomials in ${\mathbb{F}}_2\left[x\right]$, and the above factorization of $B\left(x\right)$ is its factorization into powers of distinct irreducibles. So,

$${\sigma }_{2^n}^{*}\left(B\right)=\prod _{i=1}^m\left(1+P_i{(x+1)}^{2^n{a}_i}\right).$$

By Lemma 15,

$$\prod _{i=1}^m\left(1+P_i{(x+1)}^{2^n{a}_i}\right)=\left(\prod _{i=1}^m\left(1+{P_i}^{2^n{a}_i}\right)\right)(x+1).$$

Since $\stackrel{m}{\prod _{i=1}}(1+{P_i}^{2^n{a}_i})=A^{2^n}$, it follows that

$${\sigma }_{2^n}^{*}\left(B\right)=\left(A^{2^n}\right)(x+1)={\left(A(x+1)\right)}^{2^n}=B{\left(x\right)}^{2^n}.$$

Therefore, $\overline{A}\left(x\right)$ is $2^n-$unitary perfect over ${\mathbb{F}}_2$. □

Lemma 17. 

If P is a Mersenne prime in ${\mathbb{F}}_2\left[x\right]$, then $x(x+1)|{\sigma }_k^{*}\left(P^n\right)$ for $n\in {\mathbb{N}}^{*}$, for all $k\in {\mathbb{N}}^{*}$.

Proof. 

If n is even, put $n=2^{t}u$ where u is odd and $t\in {\mathbb{N}}^{*}.$

$$\begin{array}{cc}\hfill {\sigma }_k^{*}\left(P^n\right)& =1+P^{kn}\hfill \\ \hfill & =1+P^{k{2}^{t}u}\hfill \\ \hfill & =\left(1+P^{2^t}\right)\left(1+P^{2^t}+...+{\left(P^{2^t}\right)}^{uk-1}\right)\hfill \\ \hfill & ={\left(1+P\right)}^{2^t}{\left(1+P+...+P^{uk-1}\right)}^{2^t}\hfill \end{array}$$

$1+P$ is divisible by $x(x+1)$ then $x(x+1)|{\sigma }_k^{*}\left(P^n\right).$

If n is odd, take $t=0.$ □

4. Factorization Patterns for ${\sigma }^{*}\left(x^{p^t}\right)$

In this section, we study the unitary divisor sum function ${\sigma }^{*}\left(x^{p^t}\right),$p is a prime integer. Explicit factorizations and divisibility properties of ${\sigma }^{*}\left(x^{p^t}\right)$ and related expressions for $P\in \left\{3,5,7,11\right\}$ obtained here reveal systematic appearances of Mersenne polynomials such as $T_1$ and $T_4$, which suggest a deeper underlying structure governing ${\sigma }^{*}\left(x^{p^t}\right)$. These results provide essential tools for the classification results developed later. Some results are not directly invoked in the proofs of the main classification theorem; however, they are included to emphasize recurring factorization patterns of ${\sigma }^{*}\left(x^{p^t}\right)$ and are useful in extending the classification to polynomials with larger values of $\omega \left(A\right)$.

The following lemma has a straightforward proof.

Lemma 18. 

Let$p$ be a prime integer, then

i)

${\sigma }^{*}\left(x^p\right)=(1+x)\sigma \left(x^{p-1}\right).$

ii)

${\sigma }^{*}\left({\left(1+x\right)}^p\right)=x\sigma \left({\left(1+x\right)}^{p-1}\right).$

Using Corollary 2 and Lemma 18, we get

Corollary 5. 

Let t be a positive integer and let$p$ be an odd prime, then

i)

${\sigma }^{*}\left(x^{2^{t}p}\right)={(1+x)}^{2^t}{\left(\sigma \left(x^{p-1}\right)\right)}^{2^t}.$

ii)

${\sigma }^{*}\left({\left(1+x\right)}^{2^{t}p}\right)=x^{2^t}{\left(\sigma \left({\left(1+x\right)}^{p-1}\right)\right)}^{2^t}.$

Some values of ${\sigma }^{*}\left(P\right),P$is Mersenne, are given in the below table:

Using ${\sigma }^{*}\left(T_1\right)$ and Corollary , we obtain

Lemma 19. 

Let t be a positive integer, then ${\sigma }_2^{*}\left(T_1^{2^{t-1}}\right)={(x^2+x)}^{2^t}.$

Lemma 20. 

Let t be a positive integer, then ${\sigma }_2^{*}\left(T_1^{2^{t-1}.3}\right)={(x^2+x)}^{2^t}{(1+x+x^4)}^{2^t}.$

Proof. 

The proof is done by induction. For $t=1,$ we have ${\sigma }_2^{*}\left(T_1^3\right)=1+{\left(T_1^3\right)}^2={\left(x^2+x\right)}^2{(1+x+x^4)}^2.$ Hence, the statement is true for $t=1$. Now assume ${\sigma }_2^{*}\left(T_1^{2^{t-1}.3}\right)={(x^2+x)}^{2^t}{(1+x+x^4)}^{2^t}$. Observe that

$$\begin{array}{cc}\hfill {\sigma }_2^{*}\left(T_1^{2^t.3}\right)& =1+{\left(T_1^{2^t.3}\right)}^2\hfill \\ \hfill & ={\left(1+T_1^{2^t.3}\right)}^2\hfill \\ \hfill & ={\left({\sigma }_2^{*}\left(T_1^{2^{t-1}.3}\right)\right)}^2\hfill \\ \hfill & ={(x^2+x)}^{2^{t+1}}{(1+x+x^4)}^{2^{t+1}}.\hfill \end{array}$$

The proof is now complete. □

Lemma 21. 

${\sigma }^{*}\left(x^{3^t}\right)$ and ${\sigma }^{*}\left({(1+x)}^{3^t}\right)$ are always divisible by $T_1,$$t\in {\mathbb{N}}^{*}.$

Proof. 

Consider the polynomial $x^3+1$ in ${\mathbb{F}}_2\left[x\right]$. We have

$$x^3+1=(x+1)T_1.$$

Thus,

$$x^3+1\equiv 0\left(mod{T}_1\right),$$

and so

$$x^3\equiv -1\equiv 1\left(mod{T}_1\right).$$

For $t\in N^{*}$, we have

$$x^{3^t}\equiv {\left(x^3\right)}^{3^{t-1}}\equiv {\left(1\right)}^{3^{t-1}}\equiv 1\left(mod{T}_1\right).$$

This implies that

$$x^{3^t}+1\equiv 0\left(mod{T}_1\right).$$

Therefore, $1+x^{3^t}$ is a multiple of $T_1$. The case of ${\sigma }^{*}\left({(1+x)}^{3^t}\right)$ is handled in a similar way. □

Lemma 22. 

${\sigma }^{*}\left(x^{3^t}\right)=(1+x)S_0{S}_1...S_{t-1},$$t\in {\mathbb{N}}^{*}.$

Proof. 

Since ${\sigma }^{*}\left(x^{3^t}\right)=1+x^{3^t},$ we prove by induction on t that $1+x^{3^t}=(1+x){\prod }_{k=0}^{t-1}{S}_k$. Let $P\left(t\right)$ be the statement:

$$P\left(t\right):1+x^{3^t}=(1+x)\prod _{k=0}^{t-1}{S}_k.$$

For the case $t=1,$$P\left(1\right):1+x^{3^1}=1+x^3=(1+x)S_0$. So $P\left(1\right)$ holds. Now, assume that the statement $P\left(t\right)$ is true for some integer $t\ge 1$:

$$1+x^{3^t}=(1+x)\prod _{k=0}^{t-1}{S}_k.$$

So,

$$P(t+1):1+x^{3^{t+1}}=(1+x)\prod _{k=0}^t{S}_k.$$

The right hand side of $P(t+1)$ is

$$\begin{array}{cc}\hfill (1+x)\left(\prod _{k=0}^{t-1}{S}_k\right)S_t& =(1+x^{3^t})S_t\hfill \\ \hfill & =(1+x^{3^t})(1+x^{3^t}+x^{2·3^t})\hfill \\ \hfill & =1+x^{3^{t+1}}.\hfill \end{array}$$

This is the left hand side of $P(t+1)$. Thus, the inductive step is complete. □

Corollary 6. 

Let $A=x^{3^t}{T}_1$$\in {\mathbb{F}}_2\left[x\right]$, then ${\sigma }^{*}\left(A\right)=x{(1+x)}^2{S}_0{S}_1...S_{t-1},$$t\in {\mathbb{N}}^{*}.$

Lemma 23. 

$T_4$ is a factor of ${\sigma }^{*}\left(x^{5^t}\right),$ for all $t\in {\mathbb{N}}^{*}.$

Proof. 

We have $x^5+1=(x+1)T_4.$ Then, follow the same steps as in the proof of Lemma 21. □

A mimic of the preceding two lemmas gives

Lemma 24. 

i) ${\sigma }^{*}\left(x^{7^t}\right)$ is always divisible by $T_2{T}_3,$$t\in {\mathbb{N}}^{*}.$

ii) 

${\sigma }^{*}\left(x^{{11}^t}\right)$ is always divisible by $D_0,$$t\in {\mathbb{N}}^{*}.$

The proof of the following lemma follows by induction on $t.$

Lemma 25. 

i) ${\sigma }^{*}\left(x^{5^t}\right)=(1+x)C_0{C}_1...C_{t-1}.$

ii) 

${\sigma }^{*}\left(x^{7^t}\right)=(1+x)R_0{B}_0{R}_1{B}_1...R_{t-1}{B}_{t-1}.$

iii) 

${\sigma }^{*}\left(x^{{11}^t}\right)=(1+x)D_0{D}_1...D_{t-1}.$

Corollary 7. 

i) ${\sigma }^{*}\left(x^{5^t}{T}_1\right)=x{(1+x)}^2{C}_0{C}_1...C_{t-1}.$

ii) 

${\sigma }^{*}\left(x^{7^t}{T}_1\right)=x{(1+x)}^2{R}_0{B}_0{R}_1{B}_1...R_{t-1}{B}_{t-1}.$

ii) 

${\sigma }^{*}\left(x^{{11}^t}{T}_1\right)=x{(1+x)}^2{D}_0{D}_1...D_{t-1}.$

5. Proof of Theorem 1

The proof proceeds by working on the possible values of $\omega \left(A\right)$. We first treat the case $\omega \left(A\right)=2$, then handle the more delicate case $\omega \left(A\right)=3.$ For $\omega \left(A\right)=3$, we show the exponent of the odd prime must be a power of 2 (Lemma 26) and that the irreducible polynomial must be from a specific set (Lemma 27). The final classification arises from the exponent comparison in the remaining cases (Lemmas 28–31).

5.1. Case $\omega \left(A\right)=2:$

Assume $A=P^a{Q}^b,$ where P and Q are irreducible in ${\mathbb{F}}_2\left[x\right]$ and a and b are positive integers. We begin by showing that odd unitary perfect polynomials over ${\mathbb{F}}_2$ do not exist.

We begin with a key observation. The following corollary establishes the nonexistence of odd unitary perfect polynomials over ${\mathbb{F}}_2$.

Corollary 8. 

Every non constant $2^n-$unitary perfect polynomial over ${\mathbb{F}}_2$ is divisible by $x(x+1)$.

Proof. 

Follows directly from Lemma 10 and Corollary 3. □

Consequently, both irreducible factors of A must be linear, and therefore $A=x^a{(x+1)}^b.$

Proposition 1. 

Let $A=P^a{Q}^b\in {\mathbb{F}}_2\left[x\right]$, then A is $2^n-$unitary perfect over ${\mathbb{F}}_2$ if and only if A is of the form ${(x^2+x)}^{2^t}$, for some $t\in \mathbb{N}$.

Proof. 

Sufficiency is immediate from the definition. For necessity, assume $A=x^a{(x+1)}^b$ hence we must have: $1+x^a={(x+1)}^a$ and $1+{(x+1)}^b=x^b$. Hence, $a=b=2^n$, for some $n\in \mathbb{N}$. □

The first part of Theorem 1 is done. Consequently the unitary perfect polynomials A with $\omega \left(A\right)=2$ are exactly the perfect polynomials.

5.2. Case $\omega \left(A\right)=3:$

Assume now that A has exactly three distinct irreducible factors. By Corollary 8, A over ${\mathbb{F}}_2$ must be even and has the form $x^a{(x+1)}^b{P}^c,$P is an odd irreducible polynomial. We begin by restricting the possible exponent of P.

Lemma 26. 

If $A=x^a{(x+1)}^b{P}^c$ is a $2^n-$unitary perfect polynomial over ${\mathbb{F}}_2$, then $c=2^t$ for some $t\in \mathbb{N}$.

Proof. 

Since A is a $2^n-$unitary perfect polynomial,

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

The expression $(1+x^{a·2^n})(1+{(x+1)}^{b·2^n})(1+P^{c·2^n})$ must have the factors x, $x+1$, and P. Considering the factor ${\sigma }_{2^n}^{*}\left(P^c\right)=(1+P^{c·2^n})={(1+P^c)}^{2^n}$. Assume $c=2^{t}u$, where u is odd and $t\in \mathbb{N}$. So

$${\sigma }^{*}\left(P^c\right)={\sigma }^{*}\left(P^{2^{t}u}\right)={\left({\sigma }^{*}\left(P^u\right)\right)}^{2^t}={(1+P^u)}^{2^t}$$

Since P is odd (neither divisible by x nor by $x+1$), then $1+P^u$ is also not divisible by P unless u is a multiple of the characteristic of the field, which is 2. But u is odd. So P and $1+P^u$ are coprime. Thus, the only possible prime factors of $1+P^u$ are x and $x+1$ so $1+P^u$ must be of the form $x^h{(x+1)}^k$ for some integers $h,k$.

But

$$\begin{array}{cc}\hfill 1+P^u& =(1+P)(1+P+\dots +P^{u-1})\hfill \\ \hfill & =x^h{(x+1)}^k.\hfill \end{array}$$

The two factors $(1+P)$ and $(1+P+\dots +P^{u-1})$ are coprime and since their product must be of the form $x^h{(x+1)}^k$, the prime factors of each of these two factors must also be a subset of $\{x,x+1\}.$ The only way for these two coprime polynomials to be made of factors of x and $x+1$ is for one of the factors to be equal to 1.

If $u>1$, the degree of the second factor,

$$deg(1+P+\dots +P^{u-1})=(u-1)deg\left(P\right),$$

is greater than 0, so it cannot be 1. This forces the case where the second factor is 1, which only happens when the sum has a single term. Thus $u-1=0$ or $u=1$.

This implies that $c=2^t$, which completes the proof. □

Lemma 27. 

Let $a=2^{h}r,$$b=2^{k}s$ with r and s are odd. If $A=x^a{(x+1)}^b{P}^{2^t}$ is a $2^n-$unitary perfect polynomial over ${\mathbb{F}}_2$, then $P\in \{T_1,T_4\}$.

Proof. 

Using the identities

$$1+x^a={(x+1)}^{2^h}{(1+x+\dots +x^{r-1})}^{2^h},$$

(1)

$$1+{(x+1)}^b=x^{2^k}{(1+(x+1)+\dots +{(x+1)}^{s-1})}^{2^k},$$

(2)

and

$$1+P^{2^t}={(1+P)}^{2^t}.$$

(3)

Lemma 6-i) implies that

$$1+x+\dots +x^{r-1},1+(x+1)+\dots +{(x+1)}^{s-1}\in \{1,P\}.$$

By symmetry, this leads to two possible configurations:

$$\begin{array}{c}Case1:1+x+\dots +x^{r-1}=P,s=1,\hfill \\ Case2:1+x+\dots +x^{r-1}=P=1+(x+1)+\dots +{(x+1)}^{s-1}.\hfill \end{array}$$

(4)

By Lemma 5-iii), we have: $P\in \{T_1,T_4\}$. □

Once the irreducible factor P is fixed, the equality ${\sigma }_{2^n}^{*}\left(A\right)=A^{2^n}$ rigidly determines all remaining exponents.

5.2.1. Subcase: $P=T_1$

Lemma 28. 

Let r be an odd integer. If $A=x^{2^{h}r}{(x+1)}^{2^k}{T}_1^{2^t}$ is a $2^n-$unitary perfect polynomial over ${\mathbb{F}}_2$, then $r=3,$$k=h+1$ and $t=h.$

Proof. 

From Case 1 in Lemma 27, we have $1+x+\dots +x^{r-1}=T_1$ which implies $r=3.$$A^{2^n}={\sigma }_{2^n}^{*}\left(A\right)$ yields

$$\begin{array}{cc}\hfill x^{2^h.3}{(x+1)}^{2^k}{T}_1^{2^t}& ={(x+1)}^{2^h}{T}_1^{2^h}{x}^{2^k}{\left(x(x+1)\right)}^{2^t}\hfill \\ \hfill & =x^{2^k+2^t}{(x+1)}^{2^h+2^t}{T}_1^{2^h}.\hfill \end{array}$$

(5)

Comparing the exponents and degrees in 5 gives $t=h$and $2^h+2^h=2^k.$ So, $k=h+1.$ □

5.2.2. Subcase: $P=T_4$

Lemma 29. 

Let r be an odd integer. If $A=x^{2^{h}r}{(x+1)}^{2^k}{T}_4^{2^t}$ is a $2^n-$unitary perfect polynomial over ${\mathbb{F}}_2$, then $r=5$ and $k=h+2$ and $t=h.$.

Proof. 

Here $1+x+\dots +x^{r-1}=T_4,$ so $r=5.$ Comparing exponents in $A={\sigma }^{*}\left(A\right)$ yields $t=h$and $2^h+3.2^h=2^k$, hence $k=h+2.$ □

5.2.3. Case 2:

We now treat case 2. The symmetry between the factors x and $x+1$ forces identical exponent behavior.

Lemma 30. 

Let r and s be odd integers. If $A=x^{2^{h}r}{(x+1)}^{2^{k}s}{P}^{2^t}$ is a $2^n-$unitary perfect polynomial over ${\mathbb{F}}_2$, then P$=T_1$ and $r=s=3$.

Proof. 

Case 2 in Lemma 27 implies $1+x+\dots +x^{r-1}=P=1+(x+1)+\dots +{(x+1)}^{s-1}.$

By Corollary 1, this forces P$=T_1.$Degree comparison then yields $r=s=3.$ □

Lemma 31. 

If $A=x^{2^h.3}{(x+1)}^{2^k.3}{T}_1^{2^t}$ is a $2^n-$unitary perfect polynomial over ${\mathbb{F}}_2$, then $k=h,t=h+1.$

Proof. 

A is $2^n-$unitary perfect. Under Case 2 of Lemma 27, symmetry between the factors x and $x+1$ forces $k=h.$Comparing the exponents of $T_1$ on both sides of the equality of $A^{2^n}={\sigma }_{2^n}^{*}\left(A\right)$ then gives $t=h+1.$ □

The proof of Theorem 1 is now complete.

Remark 1. 

The classification of $2^n-$unitary perfect polynomials A with a relatively large number $\omega \left(A\right)$ is likely to be highly intricate. The development of new methods will be necessary in order to achieve further progress in this direction.

6. Conclusion

We completely classified all $2^n-$unitary perfect polynomials over ${\mathbb{F}}_2$ with at most three distinct irreducible factors. We proved that no odd polynomial can satisfy the $2^n-$unitary perfect condition. The classification shows that such polynomials must have a highly restricted structure, built from powers of x, $x+1$, and a single Mersenne prime P over ${\mathbb{F}}_2$.

These results extend previous work on perfect and unitary perfect polynomials and provide a foundation for further investigations involving more irreducible factors or analogous problems over other finite fields.