On Some Polynomials Allegedly Related to the Abc Conjecture

1. Introduction The main goal of this paper is to bring your attention to the following family of polynomials. Deenition 1.1. For every a = b+c, where a, b, c are coprime natural numbers the abc-polynomial f abc (x) = bx a ? ax b + c (x ? 1) 2 : I discovered these polynomials when pursuing a rather naive approach to the Masser-Oesterl e's abc conjecture. The following argument describes the idea. It's extremely vague and I would be very happy to hear any comments on how to make it more precise or why it is doomed to fail. Argument. Although the arithmetic abc conjecture is a great mystery , its algebraic counterpart is a rather easy theorem. It looks like it was rst noticed by W. W. Stothers (cf 21]). Later on it was generalized and rediscovered independently by several people, including R.C. Mason (cf. 12]) and J. Silverman (cf. 19]). This list might be incomplete and I would appreciate any amendments to it. Theorem. Suppose a + b + c = 0; where a; b; c are coprime, not all constant, polynomials with coeecients in a eld K; charK = 0: Suppose R(x) 2 Kx] is the product of all irreducible monic polynomials from Kx] that divide abc: Then deg R max(deg a; deg b; deg c) + 1: There are several proofs of this theorem, all involving derivatives or diierential forms. I will discuss two of them, probably the easiest ones and then try to translate them into the arithmetical setting. Let's diierentiate the equality a(x) + b(x) + c(x) = 0 with respect to x: Then we get two equalities. a + b + c = 0 a 0 + b 0 + c 0 = 0 Together they imply that a b a 0 b 0 = b c b 0 c 0 = c a c 0 a 0 If we denote the above determinant by D(x); then we have the following. 1) If D(x) = 0; this would imply that ab 0 = a 0 b; so aja 0 b; so aja 0 ; so a 0 = 0 (because deg a 0 < deg a:) Therefore, because charK = 0; a(x) is a constant. As the same can be done for b and c and because we assumed that a; b; c are not all constants, we conclude that D(x) …