MASON-STOTHERS THEOREM AND PERFECT BINARY POLYNOMIALS

. We prove that there is no perfect binary polynomial R that is the sum of two appropriate powers, besides, possibly R = P +1 with P irreducible. The proofs follow from analogue results involving the ABC-theorem for polynomials and a classical identity.


Introduction
A perfect number n is a positive integer such that the sum of all its divisors including 1 and n equals 2n. For example, n = 6 and n = 28 are perfect numbers.
It is conjectured that there is an infinity of perfect numbers and that all of them are even. Deep computations have resulted in more known examples. However, essentially, we only know two main theoretical results about them, namely (a) all even perfect numbers have exactly two prime divisors, more precisely, the even perfect numbers are exactly the integers of the form 2 p−1 (2 p − 1) with 2 p − 1 prime, and (b) all, if any exist, odd perfect numbers are products of powers of primes p 4k+1 by perfect squares. Beginning with E. F. Canaday (see [2]), the first doctoral student of Leonard Carlitz, the study of an analogous problem over polynomials instead of numbers, started in 1941. More precisely, let A ∈ F 2 [x] be a binary polynomial. We say that A is perfect if and only if A equals the sum of all its divisors including 1 and A. We also say that a binary polynomial is even if B(1) = 0 or B(0) = 0, and we say that a polynomial that is not even is odd. In other words, B odd means that B(0) = 1 and B(1) = 1.
Canaday found the infinite family of even perfect polynomials (x(x + 1)) 2 n −1 where n = 0, 1, 2, . . . and called them "trivial". He also found a list of 11 non-trivial perfect polynomials, that we call sporadic (see Proposition 2.1 below), all of them even, of degrees between 5 and 20. He says furthermore, that it seems plausible that no odd perfect polynomial can exist but that this is not proved. No new perfect 2 LUIS H. GALLARDO polynomial has been discovered since besides the efforts of Beard et al. [1], Gallardo and Rahavandrainy [6,7,8,9,10], Cengiz et al. [3]. On the other hand, the polynomial ABC-theorem, i.e., Mason-Stothers theorem (see [11, pp. 194-195]), is a nice result about polynomials (see Lemma 2.4 below for details) that essentially says that if one has two coprime polynomials in one variable over a field K, besides the case where both these polynomials are p-th powers, where p is the characteristic of the field K, the degree of the product of all prime (irreducible) polynomials that divide their sum cannot be too small in terms of the degrees of both polynomials.
In the present paper, by considering possible perfect polynomials that are (essentially) sums of two powers, (generalizing the "trivial" family described above) we extend the number of cases for which we know that perfect binary polynomials cannot exist and, moreover, we propose a conjecture that seems non-trivial.
More precisely, we prove in Theorem 1.1 that we cannot build odd perfect polynomials by adding to a polynomial that splits in F 2 (e.g., a "trivial" perfect polynomial) any power of another polynomial. While our result in Theorem 1.2 characterizes the even perfect polynomials of the form P + 1, where P is irreducible, as the only perfect polynomials that are sums of two appropriate powers of binary polynomials.
Our first result is Theorem 1.1. There is no odd perfect binary polynomial R of the form R = , t > 1 is an integer, k, l are non-negative integers, and k, l are not both even.
Our second result is Assume that a perfect binary polynomial R satisfies the condition: in which P is a prime (irreducible) binary polynomial, S a binary polynomial, not divisible by P and n, k, m are non-negative integers with m ≥ 1, k ≥ 1 such that P k and S n are not both squares in F 2 [x], and one has either n = 0 and deg(R) ≤ deg(P ), or n = 0 and 1 m and gcd(m, n deg(S)) = 1.

ABC AND PERFECT BINARY POLYNOMIALS
3 Then R = P + 1.
Moreover, R is even perfect.
Our conjecture is Remark 1.4. Clearly R 0 (x) satisfies the conjecture. It is not difficult to check that of the 11 known sporadic perfect binary polynomials (see Proposition 2.1), are the only that satisfy the conjecture. While for any, say T , of the trivial even perfect binary polynomials, with deg(T ) > 2, T + 1 is always reducible (indeed, it has x 2 + x + 1 as a prime factor). Moreover, from computations in [3] it is known that the conjecture holds when deg(R) ≤ 200.
For information on the analogue of the conjecture over the integers, the reader may check [4] as well as [5].
In Section 2, one finds the necessary tools (Lemma 2.5 and Lemma 2.7) that essentially prove our main results. However, for clarity, both theorems are proved in Section 3.
A simple, but important property of binary polynomials is

LUIS H. GALLARDO
Proof. Observe that f : The following classical lemma (see [2]) is useful.
where rad(abc) is the product of all distinct prime divisors of abc in F 2 [x].
Lemma 2.5. Let k, l be non-negative integers not both even and let t > 1 be an integer. Then there are no polynomials L, M ∈ F 2 [x] such that L is odd and Proof. Assume, on the contrary, the existence of two polynomials L and M ∈ F 2 [x] satisfying (8). Put A := L 2 , B := x k (x+1) l , C := M t . Since M t cannot be a square we have t = 2t 1 + 1 and only two cases to consider.
Case 1. We have that k, l are not both odd, say k := 2k 1 , and l := 2l 1 + 1, since we can always change x by x + 1 if necessary in (8).
ABC AND PERFECT BINARY POLYNOMIALS

5
It follows from the uniqueness of T and V in (10) (guaranteed by Lemma 2.2 again) and by (11) that which is a contradiction since M is odd, so that it cannot have roots in F 2 .
Case 2. We have that k, l are both odd. We obtain the same contradiction as before, since we get the same V on the left hand side of (10). This finishes the proof of the Lemma.
Remark 2.6. We first found a proof of Lemma 2.5 using the ABC theorem, however our present proof above, based on the classical Lemma 2.2 is much shorter.
Lemma 2.7. Assume that a non-constant binary polynomial R satisfies the condition: in which P is a prime (irreducible) binary polynomial, S a binary polynomial, not divisible by P and n, k, m are non-negative integers with m ≥ 1, k ≥ 1 such that P k and S n are not both squares in F 2 [x], and one has either n = 0 and deg(R) ≤ deg(P ), or n = 0 and 1 m and gcd(m, n deg(S)) = 1.
It follows from (29) that indeed Now hypothesis (14), together with (30), implies that so that k = 1. We now put together (31), deg(A) < deg(B) and hypothesis (15) to obtain m deg(R) < deg(P ) ≤ deg(R), i.e., m < 1. This is a contradiction. Therefore, Case 3 does not happen. This finishes the proof of the Lemma. Proof of Theorem 1.2: Proof. By Lemma 2.7, one has R = P + 1. Since R is perfect we cannot have deg(P ) = 1, thus deg(P ) ≥ 2 so that P is odd. It follows that R is even perfect.