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

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16 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 ) = σ ( σ ( . . . ( σ ( k times 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 { 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 ) = 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 ) 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 ) 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 ) 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 { 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 .
  • A ¯ is the polynomial obtained from A with x replaced by x + 1 , that is 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.
If ω ( A ) 2 and A is a bi-unitary superperfect over F 2 if and only if A , A ¯ { 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 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 ) 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 F 2 [ x ] be irreducible. Then
i) If a { 4 r , 4 r + 2 } , where 2 r 1 or 2 r + 1 is of the form 2 α u 1 , u odd, then σ * * ( P a ) = ( 1 + P ) 2 α · σ ( P 2 r ) · ( 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 · ( 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 N * . If α j N , then
1)
x ( x + 1 ) divides σ * * ( M i c ) .
2)
σ * * ( M 1 c ) = x a ( x + 1 ) a Π j Q j α j .
3)
σ * * ( M 2 c ) = x a ( x + 1 ) 2 a Π j Q j α j .
4)
σ * * ( M 3 c ) = x 2 a ( x + 1 ) a Π j Q j α j .
5)
σ * * ( M 4 c ) = x a ( x + 1 ) 3 a Π j Q j α j .
6)
σ * * ( M 5 c ) = x 3 a ( x + 1 ) a Π j 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 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 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 N .
i)If T is a Mersenne prime and if T = T * , then T { 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 6 .
iv)If σ ( x 2 n ) is a Mersenne prime, then 2 n { 2 , 4 } .
Lemma 9.
[Lemma 2.6 in [14]] Let m 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 ] if 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 :
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 .
Lemma 12.
[Theorem 1.1 in [12]] Let A F 2 [ x ] be bi-unitary perfect polynomial such that ω ( A ) = 3 . Then A , A ¯ { B j : j 7 } .
Lemma 13.
[Theorem 1.2 in [12]] Let A F 2 [ x ] be bi-unitary perfect polynomial such that ω ( A ) = 4 . Then A , A ¯ { B j : 8 j 13 } { R 1 , R 2 } .
Lemma 14.
If A ( x ) is a bi-unitary superperfect polynomial over F 2 , then so is 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 + P ( 0 ) + + P 2 n + 1 ( 0 ) ( 2 n + 1 ) - times = 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 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 N * and B 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 N * .
Corollary 3.
If A F 2 [ x ] and ω ( A ) 2 , then x ( x + 1 ) divides σ * * ( A ) .
Proof. 
Let ω ( A ) 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 { x , 1 + x } and the result follows by Lemma 15. □
Corollary 4.
Let A be a polynomial in F 2 [ x ] with ω ( A ) 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 ) 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 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 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 5 , 9 , 13 , 2 d 1 for some d 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 1 . But σ * * 2 ( x a ) = σ * * ( 1 + x ) 2 α · σ ( x 2 r ) · 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 6 and hence u 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 2 d 1 and since σ * * 2 ( x a ) = x 2 α 1 · 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 6 and hence u 3 , 5 , 7 . For u = 3 , σ * * 2 ( x a ) = x 2 α 1 · σ * * σ ( x 2 ) 2 α = x 2 α 1 · σ * * M 1 2 α . Hence, α = 1 and the same result is obtained when u 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 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
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 .

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.
If A { 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.
If A = P α 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 N * and let A = P a be a bi-unitary superperfect polynomial over F 2 , then P 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 , A ¯ { x 2 , x 2 d 1 } , where d N * .
Proof. 
By Corollary 5, A = x α or x + 1 α . Assume A = x α and α = 2 m , then σ * * 2 ( A ) = σ * * x m + 1 + 1 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 . If α = 2 m + 1 , then σ * * 2 ( A ) = σ * * 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 N * . Then, σ * * 2 ( A ) = σ * * 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 A ¯ . □

4.2. Case w(A)=2

We consider the polynomial A = P a Q b and a , b 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 3 or b 3 ) and ( a 2 n 1 or b 2 m 1 for any n , m 1 ) .
Lemma 22.
Let a , b , d 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 ) { ( 9 , 9 ) , ( 9 , 13 ) , ( 13 , 9 ) } .
3)
If a and b are even, then a = b { 2 , 4 } .
4)
If a is odd and b is even, then ( a , b ) 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 3 and α = 1 ) or ( v 1 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 6 ) or ( v 1 6 ) and we have u , v 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 { 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 { 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 .
If 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 { 2 , 4 } . So, a = b = 4 . If 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 if 2 m + 1 is of the form 2 d 1 . If 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 { 2 , 4 , 9 , 13 , 2 d 1 } , d is a positive integer and a = b if and only if a , b { 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 { x a , ( x + 1 ) a , M i b } with 1 a 13 , 1 b 7 .
Preprints 87941 i001
Preprints 87941 i002

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