Squares from sums of fixed powers (Q2786989)

Let \(p\) and \(q\) be relatively prime positive integers. In this paper, it is proved that the equation \(p^x+q^x = y^2\) has at most two solutions in positive integers \(x\) and \(y\) which satisfy \(x \leq 3\). If there are two solutions \((x_1,y_1)\) and \((x_2,y_2)\), with \(x_1 < x_2\), then one of \(p\) or \(q\), say \(p\), is even and there are coprime integers \(m\) and \(n\) such that either \(x_1 = 1\), \(x_2 = 2\), \(p= 4mn(m^2-2mn+2n^2)\) and \(q = (m^2+2n^2)(m^2-4mn+2n^2)\) or \(x_1 = 1\), \(x_2 = 3\), \(p= (n-m)(3n-m)(m^2+3n^2)/4\) and \(q = mn(m^2-3mn+3n^2)\). In this latter case, we may assume that both \(m\) and \(n\) are odd. As an immediate corollary we have that in case where the integers \(p\) and \(q\) are odd, the aforementioned equation has at most a single solution which satisfies \(x\in \{1,3\}\). The proof is relied on a result of \textit{H. Darmon} and \textit{L. Merel} which reduces the problem to one with \(x\in \{1,2,3\}\) [J. Reine Angew. Math. 490, 81--100 (1997; Zbl 0976.11017)] and on Chabauty's method which is used to determine the rational points of a particular curve of genus 2.