On Even 2ⁿ-Unitary Perfect Polynomials over ${\mathbb{F}}_{\mathbf{2}}$

Abstract

Let k be a positive integer. A polynomial $A\in {\mathbb{F}}_2\left[x\right]$ is called k-unitary perfectif the sum of the k-th powers of its distinct unitary divisors equals $A^k$. In this paper, we study the case $k=2^n$ and prove that every $2^n$-unitary perfect polynomial over ${\mathbb{F}}_2$ is even. Moreover, we completely classify all even $2^n$-unitary perfect polynomials having at most three distinct irreducible factors. In particular, we characterize all such polynomials of the form $A=x^a{(x+1)}^b{P}^h$, where P is a Mersenne prime over ${\mathbb{F}}_2$ and $a,b,h\in \mathbb{N}$. As a consequence, several explicit infinite families of k-unitary perfect polynomials over ${\mathbb{F}}_2$ are obtained.

1. Introduction

The study of perfect numbers and their analogs-unitary perfect and bi-unitary perfect numbers has a long and rich history in classical number theory. A classical result of Euclid [1] 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, 87, 360, …), and the existence of infinitely many remains open. Importantly, no odd unitary perfect integer has been discovered.

Motivated by these classical problems, several authors [2,3,4,5,6,7] have investigated polynomial analogs 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. [8] 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 primeThe 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. [9] introduced a polynomial analog 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: ${\sigma }_k\left(A\right)={\displaystyle \sum _{B\mid A}}{B}^k.$ If $A={\displaystyle \prod _{i=1}^m}{P}_i^{{\alpha }_i},$ then

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

Chehade et al. [9] 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 $x^a{(x+1)}^b{\displaystyle \prod _{i=1}^m}{P}_i^{h_i}$, where each $P_i$ is Mersenne and $a,b,$ and $h_i$ are positive integers.

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, ${\sigma }_k^{*}\left(A\right)={\displaystyle \underset{gcd(B,A/B)=1}{\sum _{B|A}}}{B}^k$ and if $A={\displaystyle \prod _{i=1}^m}{P}_i^{h_i},$ then

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

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

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

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

The unitary perfect polynomial have been studied extensively by Beard and Harbin [2,3,4] and by Gallardo and Rahavandrainy [7,10].

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.

A natural question is whether the techniques developed in this paper can be extended to k-unitary perfect polynomials for arbitrary positive integers k, rather than only for $k=2^n$. Our approach relies fundamentally on the characteristic-2 structure of ${\mathbb{F}}_2$ and, in particular, on the Frobenius identity

$${(a+b)}^{2^n}=a^{2^n}+b^{2^n},$$

which permits the simplification

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

when k is a power of 2 (see Lemma 12 and Corollary 2). For arbitrary integers k, mixed terms generally appear in the binomial expansion and this property no longer holds in general. Consequently, several factorization and divisibility arguments used in the present work cannot be transferred directly.

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 analog 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\}$:

$$\begin{array}{c}{S}_t=1+x^{3^t}+x^{{2.3}^t}, \hfill \\ R_t=1+x^{7^t}+x^{{3.7}^t},\hfill \\ B_t=1+x^{{2.7}^t}+x^{{3.7}^t},\hfill \\ C_t=1+x^{5^t}+x^{{2.5}^t}+x^{{3.5}^t}+x^{{4.5}^t},\hfill \\ D_t=1+x^{{11}^t}+x^{{2.11}^t}+\dots +x^{{10.11}^t}.\hfill \end{array}$$

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

([7]). 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

([7]). 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

([7]). 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 [7], Gallardo and Rahavandrainy provided a complete classification of 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 [11] (Lemma 2)).

Lemma 5. (i)

Any complete polynomial is equal to its reciprocal polynomial.

(ii) 

If $1+x+\cdots +x^h=PQ$, 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

([11] (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 $1+P+\cdots +P^{2n}=Q^m$, then $m\in \{0,1\}$.

(ii) 

If $1+P+\cdots +P^{2n}=Q^{m}A$, where $m>1$ and $A\in {\mathbb{F}}_2\left[x\right]$ is nonconstant, then $\mathrm{deg}\left(P\right)>\mathrm{deg}\left(Q\right)$.

(iii) 

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

(iv) 

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

(v) 

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

Lemma 7.

1. 

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

2. 

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

3. 

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

As a special case of [12] (Theorem 2.47), it is obtained that

Lemma 8.

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

Consequently, it is derived that

Lemma 9.

(i) 

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

(ii) 

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

(iii) 

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

Lemma 10

([7]). Any nonconstant 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 $x(x+1).$

Proof. 

Since P is a Mersenne prime over ${\mathbb{F}}_2$, we may write

$$P=1+x^a{(x+1)}^b$$

for some positive integers a and b with $gcd(a,b)=1$. Hence

$$1+P=x^a{(x+1)}^b,$$

which is divisible by $x(x+1)$. Since

$${\sigma }^{*}\left(P^n\right)=1+P^n,$$

we have

$$1+P^n=(1+P)(1+P+\cdots +P^{n-1}).$$

Therefore, $1+P$ divides ${\sigma }^{*}\left(P^n\right)$, and, consequently,

$$x(x+1)\mid {\sigma }^{*}\left(P^n\right).$$

This proves the claim. □

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 $k=2^n$, n is a positive integer.

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 $A={\displaystyle \prod _{i=1}^r}{P}_i^{{\alpha }_i}\in {\mathbb{F}}_2\left[x\right]$ 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 2.

Corollary 3.

$A={\displaystyle \prod _{i=1}^m}{P}_i^{{\alpha }_i}$ 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. 

$$\begin{array}{cc}\hfill {\sigma }_{2^n}^{*}\left(A\right)& ={\left({\sigma }^{*}\left(A\right)\right)}^{2^n}\hfill \\ \hfill \hspace{1em}& ={\left({\sigma }^{*}\left(A\right)\right)}^{2^m}{\left({\sigma }^{*}\left(A\right)\right)}^{2^{n-m}}\hfill \\ \hfill \hspace{1em}& ={\sigma }_{2^m}^{*}\left(A\right){\left({\sigma }^{*}\left(A\right)\right)}^{2^{n-m}}.\hfill \end{array}$$

□

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 $A\mid {\sigma }_k^{*}\left(A\right)$. Then there exists $Q\in {\mathbb{F}}_2\left[x\right]$ such that ${\sigma }_k^{*}\left(A\right)=QA$.

Since $A=P^{\alpha }$, we have

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

Hence

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

Therefore,

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

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 14.

As a result of Corollary 14 and Lemma 4, 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

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

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,

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

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

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

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

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

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,

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

By Lemma 15,

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

Since ${\displaystyle \prod _{i=1}^m}(1+{P_i}^{2^n{a}_i})=A^{2^n}$, 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 $t\in {\mathbb{N}}^{*}$, then

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

Since $1+P$ 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 ${\mathit{\sigma }}^{\mathbf{*}}\left({\mathit{x}}^{{\mathit{p}}^{\mathit{t}}}\right)$

In this section, we study the unitary divisor sum function ${\sigma }^{*}\left(x^{p^t}\right),$ where 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 ${\sigma }^{*}\left(P\right)$, where P is a Mersenne polynomial over ${\mathbb{F}}_2$, are summarized in

Table 1. Some values of ${\sigma }^{*}\left(P\right)$.

P	${\mathit{\sigma }}^{*}\left(\mathit{P}\right)$
$T_1$	$x(1+x),$
$T_2$	$x{(1+x)}^2,$
$T_3$	$x^2(1+x),$
$T_4$	$x{(1+x)}^3,$
$T_5$	$x^3(1+x).$

.

Using ${\sigma }^{*}\left(T_1\right)$ and Corollary 2, 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

${\sigma }_2^{*}\left(T_1^{{3.2}^{t-1}}\right)={(x^2+x)}^{2^t}{(1+x+x^4)}^{2^t}.$

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 ${\sigma }_2^{*}\left(T_1^{{3.2}^{t-1}}\right)={(x^2+x)}^{2^t}{(1+x+x^4)}^{2^t}$. Observe that

$$\begin{array}{cc}\hfill {\sigma }_2^{*}\left(T_1^{{3.2}^t}\right)& =1+{\left(T_1^{2^t.3}\right)}^2\hfill \\ \hfill \hspace{1em}& ={\left(1+T_1^{{3.2}^t}\right)}^2\hfill \\ \hfill \hspace{1em}& ={\left({\sigma }_2^{*}\left(T_1^{{3.2}^{t-1}}\right)\right)}^2\hfill \\ \hfill \hspace{1em}& ={(x^2+x)}^{2^{t+1}}{(1+x+x^4)}^{2^{t+1}}.\hfill \end{array}$$

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,

$$x^3+1\equiv 0 (mod T_1),$$

and so

$$x^3\equiv -1\equiv 1 (mod T_1).$$

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

$$x^{3^t}\equiv {\left(x^3\right)}^{3^{t-1}}\equiv {\left(1\right)}^{3^{t-1}}\equiv 1 (mod T_1).$$

This implies that

$$x^{3^t}+1\equiv 0 (mod T_1).$$

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.

Let $t\in {\mathbb{N}}^{*}$. Then

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

Proof. 

Since

$${\sigma }^{*}\left(x^{3^t}\right)=1+x^{3^t},$$

it suffices to prove that

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

by induction on t.

For $t=1$, we have

$$1+x^3=(1+x)(1+x+x^2)=(1+x)S_0.$$

Thus the statement holds.

Assume that for some $t\ge 1$,

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

Then

$$(1+x){\displaystyle \prod _{k=0}^t}{S}_k=\left((1+x){\displaystyle \prod _{k=0}^{t-1}}{S}_k\right)S_t$$

and by the induction hypothesis,

$$=(1+x^{3^t})\left(1+x^{3^t}+x^{2·3^t}\right).$$

Expanding in characteristic 2 gives

$$(1+x^{3^t})(1+x^{3^t}+x^{2·3^t})=1+x^{3^{t+1}}.$$

Hence

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

Therefore the result follows by induction. □

Corollary 6.

Let $t\in {\mathbb{N}}^{*}$ and let $A=x^{3^t}{T}_1$$\in {\mathbb{F}}_2\left[x\right]$. Then

$${\sigma }^{*}\left(x^{3^t}\right)=x{(1+x)}^2{\displaystyle \prod _{k=0}^{t-1}}{S}_k.$$

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. □

Using arguments analogous to those in the preceding two lemmas, we obtain

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) 

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

(ii) 

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

(iii) 

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

Corollary 7.

(i) 

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

(ii) 

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

(iii) 

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

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

Proof. 

This 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 $n\in \mathbb{N}$. □

This completes the proof of the first part of Theorem 1. 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,

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

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. Consider 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

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

The two factors $(1+P)$ and $(1+P+\cdots +P^{u-1})$ 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,

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

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

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

(1)

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

(2)

and

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

(3)

Lemma 6-(i) implies that

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

By symmetry, this leads to two possible configurations:

$$\begin{array}{c}Case 1:1+x+\cdots +x^{r-1}=P, s=1,\hfill \\ Case 2:1+x+\cdots +x^{r-1}=P=1+(x+1)+\cdots +{(x+1)}^{s-1}.\hfill \end{array}$$

(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 $1+x+\cdots +x^{r-1}=T_1$ which implies $r=3.$ $A^{2^n}={\sigma }_{2^n}^{*}\left(A\right)$ yields

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

(5)

Comparing the exponents and degrees in Equation (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 $1+x+\cdots +x^{r-1}=T_4,$ 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 $1+x+\cdots +x^{r-1}=P=1+(x+1)+\cdots +{(x+1)}^{s-1}.$

By Corollary 1, this forces P $=T_1.$ Degree comparison then yields $r=s=3.\hspace{1em}$□

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. Conclusions

We completely classified all $2^n$-unitary perfect polynomials over ${\mathbb{F}}_2$ with at most three distinct irreducible factors and proved that no odd polynomial satisfies the $2^n$-unitary perfect condition. The classification reveals a highly restricted structure involving powers of x, $x+1$, and a single Mersenne prime. The present classification was established specifically over the polynomial ring ${\mathbb{F}}_2\left[x\right]$, where the characteristic-2 structure plays an essential role in simplifying the behavior of the k-unitary divisor function and in deriving the factorization properties used throughout the proofs. In particular, several arguments rely on identities that become especially tractable in characteristic 2.

Future work may consider extending the classification to polynomials with more than three irreducible factors and exploring analogous problems over more general finite fields ${\mathbb{F}}_q\left[x\right]$, where $q>2$. Such extensions may lead to new structural insights into k-unitary perfect polynomials and require techniques beyond those used in the present work.