Problem
NT-B2-M08-P012 Fourth Powers and a Square
Prove that the equation \(x^4+y^4=7z^2\) has no solutions in positive integers.
First examine fourth powers modulo \(7\), then track how strongly \(z\) is divisible by \(7\).
Fourth powers modulo \(7\) are \(0,1,2,4\). A sum of two such residues can be \(0\pmod 7\) only as \(0+0\). Hence \(7\mid x\) and \(7\mid y\). Write \(x=7x_1\), \(y=7y_1\). Then the left side is divisible by \(7^4\), so \(7z^2\) is divisible by \(7^4\), meaning \(z^2\) is divisible by \(7^3\). Since valuations in a square are even, \(7^2\mid z\); write \(z=49z_1\). Substitution gives \(7^4(x_1^4+y_1^4)=7\cdot 7^4 z_1^2\), hence \(x_1^4+y_1^4=7z_1^2\). We have obtained a smaller positive solution. Infinite descent is impossible, so no original solution exists.
This problem is inspired by the local classical number theory collection; the wording and solution are prepared as an original training version.