Problem
NT-B2-M08-P015 The General Case \(p\equiv 3\pmod 4\)
#15
★★★★★ Level 5 of 5
Let \(p\equiv 3\pmod 4\) be prime. Prove that \(x^2+y^2=pz^2\) has no nonzero integer solutions.
First prove that \(p\mid x,y\), then that \(p\mid z\).
The equation gives \(p\mid x^2+y^2\). By the lemma for primes \(3\pmod 4\), \(p\mid x\) and \(p\mid y\). Then \(p^2\mid x^2+y^2=pz^2\), so \(p\mid z^2\), hence \(p\mid z\). Dividing \(x,y,z\) by \(p\), we get a smaller nonzero solution of the same equation. This gives infinite descent, impossible in positive integers.
An important template for stronger sum-of-squares problems.