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

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

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Submitted:

16 October 2023

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17 October 2023

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Abstract

In this paper, we give all non splitting bi-unitary superperfect polynomials divisible by one or two irreducible polynomials over the prime field of two elements. We prove the nonexistence of odd bi-unitary superperfect polynomials over F2.

Keywords:

Subject: Computer Science and Mathematics - Algebra and Number Theory

1. Introduction

Let n and k be positive integers and let

σ (n)

(resp.

σ^{*} (n)

) 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

σ^{k} (n) = \underset{k - times}{\underset{︸}{σ (σ (. . . (σ (}} n)))) = 2 n

. 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

σ^{* 2} (n) = σ^{*} (σ^{*} (n)) = 2 n .

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

σ^{* *} (n)

denotes the sum of positive bi-unitary divisors of the integer n. Wall [3] proved that there are only three bi-unitary perfect numbers

(σ^{* *} (n) = 2 n)

, namely 6, 60 and 90. Yamada [4] proved that 2 and 9 are the only bi-unitary superperfect numbers, that is

σ^{* * 2} (n) = 2 n

if and only if

n \in {2, 9}

Now, let A be a nonzero polynomial defined over the prime field

F_{2}

. A divisor B of A is unitary (resp. bi-unitary) if

g c d (B, A / B) = 1

(resp.

g c d_{u} (B, A / B) = 1)

, where

g c d_{u} (A, A / B)

denotes the greatest common unitary divisor of B and

A / B

. We denote by

σ

the sum of the monic divisors B of

A,

that is,

σ (A) = \sum_{B ∣ A} B .

σ^{*} (A)

(resp.

σ^{* *} (A))

represents the sum of all unitary (resp. bi-unitary) monic divisors of

A .

Note that all the functions

σ,

σ^{*}

and

σ^{* *}

are multiplicative and degree preserving.

A is an even polynomial if it has a linear factor in

F_{2} [x]

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

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

ω (A)

denotes the number of distinct irreducible monic polynomials that divide A. The notion of perfect polynomials over

F_{2}

was introduced first by Canaday [5]. A polynomial A is perfect if

σ (A) = A .

Canaday studied the case of even perfect polynomials with

ω (A) \leq 3

. In the past few years, Gallardo and Rahavandrainy [6,7,8] showed the non-existence of odd perfect polynomials over

F_{2}

with either

ω (A) = 3

or with

ω (A) \leq 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

σ^{*} (A) = A

(resp.

σ^{* *} (A) = A

). Also, A is called a unitary (resp. a bi-unitary) superperfect if

σ^{* 2} (A) = σ^{*} (σ^{*} (A)) = A

(resp.

σ^{* * 2} (A) = σ^{* *} (σ^{* *} (A)) = A)

Note that the function

σ^{* * 2}

is degree preserving but not multiplicative, and this is the main challenge in this work. So, working on bi-unitary superperfect polynomial over

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

F_{2}

. The authors in [7,9,10] list the unitary perfect polynomials over

F_{2}

with

ω (A) \leq 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

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

σ^{*} (A)

Notations: We use the following notations throughout the article.

$N$ (resp. $N^{*})$ represents the set of non-negative (resp. positive) integers.
$d e g (A)$ denotes the degree of the polynomial $A .$
$\bar{A}$ is the polynomial obtained from A with x replaced by $x + 1$ , that is $\bar{A} (x) = A (x + 1)$ .
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

F_{2}

that are divisible by at most two distinct irreducible factors, (Theorem 1). Bi-unitary superperfect polynomials over

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.

ω (A) \leq 2

and A is a bi-unitary superperfect over

F_{2}

if and only if

A, \bar{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 N^{*} .

2. Preliminaries

The following two lemmas are helpful.

Lemma 1.

Let A be a polynomial in

F_{2} [x]

, then

σ^{*} (A^{2^{n}}) = {(σ^{*} (A))}^{2^{n}},

n is a non-negative integer.

Proof.

The result follows since

σ^{*}

is multiplicative and

σ^{*} (p^{2^{n}}) = 1 + p^{2^{n}} = {(1 + p)}^{2^{n}} = {(σ^{*} (p))}^{2^{n}}

. □

Lemma 2.

If A is a unitary superperfect polynomial over

F_{2}

, then

A^{2^{n}}

is also a unitary superperfect polynomial over

F_{2}

for all non-negative integers n.

Proof.

Let A be a unitary superperfect and let

B = σ^{*} (A)

. By Lemma 1, we have

σ^{* 2} (A^{2^{n}}) = σ^{*} (σ^{*} (A^{2^{n}})) = σ^{*} (B^{2^{n}}) = {(σ^{*} (B))}^{2^{n}} = {(σ^{*} (σ^{*} (A)))}^{2^{n}} = A^{2^{n}}

. □

Lemma 3.

[Lemma 2.4 in [13]] Let A be a polynomial in

F_{2} [x] .

1): If P is an odd prime factor of $A,$ then $x (x + 1)$ divides $σ^{*} (A) .$
2): If $x (x + 1)$ divides $A,$ then $x (x + 1)$ divides $σ^{*} (A) .$
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

ω (A)

\geq 2 .

Lemma 5.

[Lemma 2.2 in [12]]

1): $σ^{* *} (P^{2 a + 1}) = σ (P^{2 a + 1}) .$
2): $σ^{* *} (P^{2 a}) = (1 + P^{a + 1}) σ (P^{a - 1}) = (1 + P) σ (P^{a}) σ (P^{a - 1}) .$

Corollary 1.

[Corollary 2.3 in [12]] Let

T \in F_{2} [x]

be irreducible. Then

i) If

a \in {4 r, 4 r + 2}

, where

2 r - 1

2 r + 1

is of the form

2^{α} u - 1

, u odd, then

σ^{* *} (P^{a}) = {(1 + P)}^{2^{α}} \cdot σ (P^{2 r}) \cdot {(σ (P^{u - 1}))}^{2^{α}}, gcd (σ (P^{2 r}), σ (P^{u - 1})) = 1 .

ii) If

a = 2^{α} u - 1

is odd, with u odd, then

σ^{* *} (P^{a}) = {(1 + P)}^{2^{α} - 1} \cdot {(σ (P^{u - 1}))}^{2^{α}}

The proof of the below lemma follows from Lemma 5 and the binomial formula. Table Section 6 shows some values of

σ^{* *} (A)

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

F_{2}

and let

a, c \in N^{*} .

α_{j} \in N,

then

1): $x (x + 1)$ divides $σ^{* *} (M_{i}^{c}) .$
2): $σ^{* *} (M_{1}^{c}) = x^{a} {(x + 1)}^{a} \underset{j}{Π} Q_{j}^{α_{j}} .$
3): $σ^{* *} (M_{2}^{c}) = x^{a} {(x + 1)}^{2 a} \underset{j}{Π} Q_{j}^{α_{j}} .$
4): $σ^{* *} (M_{3}^{c}) = x^{2 a} {(x + 1)}^{a} \underset{j}{Π} Q_{j}^{α_{j}} .$
5): $σ^{* *} (M_{4}^{c}) = x^{a} {(x + 1)}^{3 a} \underset{j}{Π} Q_{j}^{α_{j}} .$
6): $σ^{* *} (M_{5}^{c}) = x^{3 a} {(x + 1)}^{a} \underset{j}{Π} Q_{j}^{α_{j}} .$

Lemma 7.

[Corollary 2.4 in [12]]

1): $σ^{* *} (x^{a})$ splits over $F_{2}$ if and only if $a = 2$ or $a = 2^{d} - 1$ , for some $d \in N^{*}$ .
2): $σ^{* *} (P^{c})$ splits over $F_{2}$ if and only if P is Mersenne and $c = 2$ or $c = 2^{d} - 1$ for some $d \in N^{*}$ .

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

Lemma 8.

Let T be irreducible in

F_{2} [x]

and let

n, m \in N

i)If T is a Mersenne prime and if

T = T^{*}

, then

T \in {M_{1}, M_{4}}

ii)If

σ (x^{2 n}) = P Q

and

P = σ ({(x + 1)}^{2 m})

, then

2 n = 8

2 m = 2

P = M_{1}

and

Q = P (x^{3}) = 1 + x^{3} + x^{6}

iii)If any irreducible factor of

σ (x^{2 n})

is a Mersenne prime, then

2 n \leq 6

iv)If

σ (x^{2 n})

is a Mersenne prime, then

2 n \in {2, 4}

Lemma 9.

[Lemma 2.6 in [14]] Let

m \in N^{*}

and M be a Mersenne prime. Then,

σ (x^{2 m})

σ ({(x + 1)}^{2 m})

and

σ (M^{2 m})

are all odd and squarefree.

3. Bi-unitary superperfect Polynomials

Recall that A is a bi-unitary superperfect polynomial in

F_{2} [x]

σ^{* * 2} (A) = σ^{* *} (σ^{* *} (A)) = A .

The polynomial

A = x^{4} {(1 + x)}^{4}

is a bi-unitary superperfect polynomial over

F_{2}

The proof of the following lemmas follow directly.

Lemma 10.

If A is a bi-unitary perfect polynomial over

F_{2}

, then A is also a bi-unitary superperfect polynomial.

Lemma 11.

If A is a bi-unitary superperfect polynomial over

F_{2}

, then

B = σ^{* *} (A)

is also a bi-unitary superperfect polynomial.

Rahavandrainy (Lemma 2.6 in [12]) proved that if A is a bi-unitary perfect polynomial over

F_{2}

where

A = A_{1} A_{2}

such that

g c d (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

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

F_{2} :

\begin{matrix} C = 1 + x + x^{4}, & B_{1} = x^{3} {(x + 1)}^{4} M_{1}, & B_{2} = x^{3} {(x + 1)}^{5} M_{1}^{2}, \\ B_{3} = x^{4} {(x + 1)}^{4} M_{1}^{2}, & B_{4} = x^{6} {(x + 1)}^{6} M_{1}^{2}, & B_{5} = x^{4} {(x + 1)}^{5} M_{1}^{3}, \\ B_{6} = x^{7} {(x + 1)}^{8} M_{5}, & B_{7} = x^{7} {(x + 1)}^{9} M_{5}^{2}, & B_{8} = x^{8} {(x + 1)}^{8} M_{4} M_{5}, \\ B_{9} = x^{8} {(x + 1)}^{9} M_{4} M_{5}^{2}, & B_{10} = x^{7} {(x + 1)}^{10} M_{1}^{2} M_{5}, & B_{11} = x^{7} {(x + 1)}^{13} M_{2}^{2} M_{3}^{2}, \\ B_{12} = x^{9} {(x + 1)}^{9} M_{4}^{2} M_{5}^{2}, & B_{13} = x^{14} {(x + 1)}^{14} M_{2}^{2} M_{3}^{2}, & R_{1} = x^{4} {(x + 1)}^{5} M_{1}^{4} C, \\ R_{2} = x^{4} {(x + 1)}^{5} M_{1}^{5} C^{2} . \end{matrix}

Lemma 12.

[Theorem 1.1 in [12]] Let

A \in F_{2} [x]

be bi-unitary perfect polynomial such that

ω (A) = 3

. Then

A, \bar{A} \in {B_{j} : j \leq 7} .

Lemma 13.

[Theorem 1.2 in [12]] Let

A \in F_{2} [x]

be bi-unitary perfect polynomial such that

ω (A) = 4

. Then

A, \bar{A} \in {B_{j} : 8 \leq j \leq 13} ⋃ {R_{1}, R_{2}} .

Lemma 14.

A (x)

is a bi-unitary superperfect polynomial over

F_{2}

, then so is

\bar{A} (x)

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 $σ^{* *} (x^{a})$ .
2): x divides $σ^{* *} ({(1 + x)}^{a})$ .

Proof.

An immediate result of Lemma 5. □

Lemma 15.

x (x + 1)

divides

σ^{* *} (P^{a})

, a is a positive integer.

Proof.

Since P is odd, then

P (0) = P (1) = 1

. If

a = 2 n + 1

, then

σ^{* *} (P^{2 n + 1}) (0) = 1 + \underset{(2 n + 1) - times}{\underset{︸}{P (0) + \dots + P^{2 n + 1} (0)}} = 1 + 2 n + 1 = 0

. If

a = 2 n

, then

1 + P^{n + 1} (0) = 0 .

So, x divides

σ^{* *} (P^{a})

for every a

\in N

. Similarly,

x + 1

divides

σ^{* *} (P^{a})

. Hence,

x (x + 1)

divides

σ^{* *} (P^{a})

. □

Lemma 16.

Let A be a polynomial in

F_{2} [x]

1): If P is an odd prime factor of $A,$ then $x (x + 1)$ divides $σ^{* *} (A) .$
2): If $x (x + 1)$ divides $A,$ then $x (x + 1)$ divides $σ^{* *} (A) .$

Proof.

1): We write $A = P^{a} B$ where $a \in N^{*}$ and $B \in F_{2} [x]$ such that $g c d (P, B) = 1 .$ But, $1 + P$ divides $σ^{* *} (A)$ 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 $a, b \in N^{*}$ .

□

Corollary 3.

A \in F_{2} [x]

and

ω (A)

\geq 2

, then

x (x + 1)

divides

σ^{* *} (A) .

Proof.

Let

ω (A)

\geq 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

F_{2} [x]

with

ω (A) \geq 2

. If A is a bi-unitary superperfect, then

x (x + 1)

divides

A .

Proof.

Let

A = σ^{* * 2} (A) = σ^{* *} (B), where B = σ^{* *} (A)

. Since

ω (A) \geq 2

, then either P or

x (x + 1)

divides A. In both cases,

x (x + 1)

divides

σ^{* *} (A) = B

(Lemma 16). So,

x (x + 1)

divides

σ^{* *} (B) = σ^{* * 2} (A)

. □

The following lemma is similar to Lemma 7.

Lemma 17.

Let

a, b \in N^{*}

, then

1): If a is even, then $σ^{* * 2} (x^{a})$ and $σ^{* * 2} ({(x + 1)}^{a})$ splits over $F_{2}$ if and only if $a \in \{2, 4, 10, 12\} .$
2): If a is odd, then $σ^{* * 2} (x^{a})$ and $σ^{* * 2} ({(x + 1)}^{a})$ splits over $F_{2}$ if and only if $a \in \{5, 9, 13, 2^{d} - 1\}$ for some $d \in N^{*}$ .

Proof.

1): If $σ^{* *} (x^{a})$ splits, the $a = 2$ (Lemma 7) and $σ^{* * 2} (x^{a}) = {(x + 1)}^{2}$ . Suppose, $σ^{* *} (x^{a})$ does not split with $a = 4 r, 2 r - 1 = 2^{α} u - 1$ , (resp. $a = 4 r + 2, 2 r + 1 = 2^{α} u - 1$ ), u is odd, $r \geq 1$ . But $σ^{* * 2} (x^{a}) = σ^{* *} ({(1 + x)}^{2^{α}} \cdot σ (x^{2 r}) \cdot {(σ (x^{u - 1}))}^{2^{α}})$ , so $σ^{* *} ({(1 + x)}^{2^{α}})$ must split. Hence, $α = 1$ and since $σ (x^{2 r})$ is odd and square free (Lemma 9), then $σ (x^{2 r})$ has a Mersenne factor. So, $2 r \leq 6$ and hence $u \leq 3$ .
2): Assume $a = 2^{α} u - 1$ , with u is odd. If $σ^{* *} (x^{a})$ splits, then $a = 2^{d} - 1$ , d is positive (Lemma 7). If $σ^{* *} (x^{a})$ does not split, then $a \neq 2^{d} - 1$ and since $σ^{* * 2} (x^{a}) = x^{2^{α} - 1} \cdot σ^{* *} ({(σ (x^{u - 1}))}^{2^{α}})$ splits, $u > 1$ . Again, by Lemma 9, $σ (x^{2 r})$ has a Mersenne factor. So, $u - 1 \leq 6$ and hence $u \in \{3, 5, 7\} .$ For $u = 3$ , $σ^{* * 2} (x^{a}) = x^{2^{α} - 1} \cdot σ^{* *} ({(σ (x^{2}))}^{2^{α}}) = x^{2^{α} - 1} \cdot σ^{* *} (M_{1}^{2^{α}})$ . Hence, $α = 1$ and the same result is obtained when $u \in \{5, 7\}$ .

The same proof is done for

σ^{* * 2} ({(x + 1)}^{a})

and the proof is compete. □

Lemma 18.

Let a and

b

have the form

2^{n} - 1

where

n \in N^{*}

and let the polynomial

A = 1 + x^{a} {(x + 1)}^{b}

be Mersenne prime over

F_{2},

then

σ^{* * 2} (A) = x^{b} {(x + 1)}^{a} .

Proof.

Let

a = 2^{n_{1}} - 1

and

b = 2^{n_{2}} - 1,

then

\begin{matrix} σ^{* * 2} (A) & = σ^{* * 2} (1 + x^{a} {(x + 1)}^{b}) \\ = σ^{* *} (σ (1 + x^{a} {(x + 1)}^{b}) \\ = σ^{* *} (x^{a} {(x + 1)}^{b}) \\ = x^{b} {(x + 1)}^{a} . \end{matrix}

□

4.1. Case w(A)=1

We prove that

σ^{* *} (A)

can not have more than one prime factor when A is a prime power.

Lemma 19.

A \in {x, x + 1}

and

σ^{* * 2} (A^{a})

splits over

F_{2}

, then A is a bi-unitary superperfect polynomial.

Proof.

Follows from part 1) of Lemma 17. □

Lemma 20.

A = P^{α} \in F_{2} [x]

, 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

σ^{* *} (A)

and by Lemma 16 we have

x (x + 1)

divides

σ^{* * 2} (A) = P^{a}

. A contradiction. □

In particular, if M is a Mersenne prime polynomial over

F_{2}

, then

M^{c}

(c is a positive integer) is never a bi-unitary superperfect polynomial.

Corollary 5.

Let

a \in N^{*}

and let

A = P^{a}

be a bi-unitary superperfect polynomial over

F_{2},

then

P \in \{x, x + 1\}

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

F_{2}

with

ω (A) = 1,

then A is a bi-unitary superperfect polynomial if and only if

A, \bar{A} \in {x^{2}, x^{2^{d} - 1}},

where

d \in N^{*} .

Proof.

By Corollary 5,

A = x^{α}

{(x + 1)}^{α} .

Assume

A = x^{α}

and

α = 2 m

, then

σ^{* * 2} (A) = σ^{* *} ((x^{m + 1} + 1) \frac{x^{m} - 1}{x - 1}) .

Both

x^{m + 1} + 1

and

x^{m} + 1

split over

F_{2}

only when

m = 1 .

Thus,

σ^{* * 2} (A) = σ^{* *} (x^{2} + 1) = x^{2} .

α = 2 m + 1

, then

σ^{* * 2} (A) = σ^{* *} (\frac{x^{2 (m + 1)} - 1}{x - 1})

. The expression

x^{2 (m + 1)} + 1

splits over

F_{2}

when

2 m + 2 = 2^{d},

d \in N^{*}

. Then,

σ^{* * 2} (A) = σ^{* *} (\frac{x^{2^{d}} - 1}{x - 1}) = 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

\bar{A}

. □

4.2. Case w(A)=2

We consider the polynomial

A = P^{a} Q^{b}

and

a, b \in N^{*} .

Note that

A = x^{2} {(1 + x)}^{2}

and

A = x^{2^{α} - 1} {(1 + x)}^{2^{α} - 1}

are bi-unitary superperfect polynomials over

F_{2}

, see Lemma 10 and (Theorem 5 in [11]

) .

Corollary 6.

If A is a bi-unitary superperfect polynomial over

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

σ^{* *} (x^{a} {(x + 1)}^{b})

does not split, then

(a \geq 3

b \geq 3)

and

(a \neq 2^{n} - 1

b \neq 2^{m} - 1

for any

n, m \geq 1)

Lemma 22.

Let

a, b, d \in N^{*} .

The polynomial

A = x^{a} {(x + 1)}^{b}

is a bi-unitary superperfect over

F_{2}

if and only if one of the following is true.

1): If a and b are odd and $σ^{* *} (x^{a} {(x + 1)}^{b})$ splits, then a and b are of the form $2^{d} - 1 .$
2): If a and b are odd and $σ^{* *} (x^{a} {(x + 1)}^{b})$ doe not split, then $(a, b) \in {(9, 9), (9, 13), (13, 9)} .$
3): If a and b are even, then $a = b \in {2, 4} .$
4): If a is odd and b is even, then $(a, b) \in \{(2, 2^{d} - 1), (2^{d} - 1, 2)\} .$

Proof.

1) If

a = 2 m + 1

and

b = 2 n + 1

, then

σ^{* * 2} (A) = σ^{* *} (σ^{* *} (x^{a}) {(1 + x)}^{b}) .

But

σ^{* *} (x^{2 m + 1})

and

σ^{* *} {(x + 1)}^{2 n + 1}

split over

F_{2}

when

2 m + 1

and

2 n + 1

are of the form

2^{d} - 1

(Lemma 7).

2) If

a = 2^{α} u - 1

and

b = 2^{β} v - 1

u, v

are odd. We have

u > 1

and

v > 1

, since

σ^{* *} (x^{a} {(x + 1)}^{b})

does not split.

σ^{* *} (x^{a} {(x + 1)}^{b}) = σ^{* *} ({(1 + x)}^{2^{α} - 1} {(σ (x^{u - 1}))}^{2^{α}} x^{2^{β} - 1} σ {({(x + 1)}^{v - 1})}^{2^{β}}) .

By Lemma 21

(u - 1 \geq 3

and

α = 1)

(v - 1 \geq 3

and

β = 1)

. Also,

σ (x^{u - 1})

and

σ ({(x + 1)}^{v - 1})

does not split since

σ^{* *} (x^{a} {(x + 1)}^{b})

does not split. So, there exit Merssene primes M(resp.

M^{'})

that divides

σ (x^{u - 1})

(resp.

σ ({(x + 1)}^{v - 1})

. Hence,

(u - 1 \leq 6)

(v - 1 \leq 6)

and we have

u, v \in \{5, 7\}

. 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 {4 r, 4 r + 2}

such that

2 r - 1

2 r + 1

is of the form

2^{α} u - 1

with u is odd and

b \in {4 r^{'}, 4 r^{'} + 2}

such that

2 r^{'} - 1

2 r^{'} + 1

is of the form

2^{β} v - 1

, v odd. Thus,

σ^{* *} (A) = {(1 + x)}^{2^{α} - 1} σ (x^{2 r}) {(σ (x^{u - 1}))}^{2^{α}} x^{2^{β} - 1} σ ({(x + 1)}^{2 r'}) {(σ ({(x + 1)}^{v - 1}))}^{2^{β}} .

σ (x^{2 r}),

σ ({(x + 1)}^{2 r'}),

σ (x^{u - 1})

and

σ ({(x + 1)}^{v - 1})

are Mersenne, then

2 r, 2 r^{'}, u - 1, v - 1 \in {2, 4} .

So,

a = b = 4 .

σ (x^{2 r}),

σ (x^{u - 1}),

σ ({(x + 1)}^{2 r'})

and

σ ({(x + 1)}^{v - 1})

are not Mersenne, then

r, r^{'}, u - 1, v - 1 > 2

and

ω (σ^{* * 2} (A)) > 2,

a contradiction. For

a = b = 2,

A is bi-unitary perfect and hence A is a bi-unitary superperfect.

4) Now, let

a = 2 m + 1

and

b = 2 n

. Since

σ^{* *} ({(x + 1)}^{2 n})

splits over

F_{2}

only when

n = 1

, then

σ^{* * 2} (A) = σ^{* *} (σ^{* *} (x^{2 m + 1}) σ^{* *} ({(x + 1)}^{2})) .

But

σ^{* *} (x^{2 m + 1})

splits over

F_{2}

2 m + 1

is of the form

2^{d} - 1 .

a = 2 m

and

b = 2 n + 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

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

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

σ^{* *} (A)

and

σ^{* * 2} (A)

for

A \in {x^{a}, {(x + 1)}^{a}, M_{i}^{b}}

with

1 \leq a \leq 13

1 \leq b \leq 7

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

Abstract

1. Introduction

2. Preliminaries

3. Bi-unitary superperfect Polynomials

4. Proof of Theorem 1

4.1. Case w(A)=1

4.2. Case w(A)=2

5. Conclusion

6. Table

References

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