The Diophantine equation 𝑓(𝑥)=𝑔(𝑦)

Abstract

Abstract.Let f(x),g(y) be polynomials over Z of degrees n and m respectively and with leading coefficients an , b . Suppose that m\n and that an/bm is the mth power of a rational number. We give two elementary proofs that the equation f(x) = g(y) has at most finitely many integral solutions unless f(x) = g(h(x)) for some polynomial h(x) with rational coefficients taking integral values at infinitely many integers.The problem of showing that a given polynomial diophantine equation (1) P(x,y) = 0 has at most finitely many integral solutions has been a central problem of research in number theory. Deep results in algebraic geometry and diophantine approximation have been employed to address this problem; (see [2,3] for surveys). For instance, in 1929 Siegel [5] used such methods to prove one of the most celebrated results in this area; namely, that if the curve given by ( 1 ) is irreducible and of positive genus then it has at most finitely many points with integral coordinates. Moreover, he was able to characterize those curves of genus zero having infinitely many points with integral coordinates. The object of this paper is to obtain the same type of result for a special class of diophantine equations as an easy corollary of two lemmas from elementary complex analysis. A second proof of the result is given at the end of the paper using only elementary methods.We shall restrict our attention to diophantine equations of the type (2) f(x) = anx +an_xx + .-+ a0 = bmy +bm_xy +---+ b0 = g(y)with integer coefficients, an / 0, bm / 0, although the method we use may be applied to a wider variety of equations of type ( 1 ).Theorem. Suppose that m\n and that (an/bm) is the mth power of a rational number. Then either (iCC) f(x) -g(h(x)) for some polynomial h(x) with rational coefficients taking integral values at infinitely many integers ; or (ii) equation (2) has at most finitely many integral solutions, u