Problem
NT-B2-M02-P020 General Descent for \(p\equiv3\pmod4\)
#20
★★★★★ Level 5 of 5
Let \(p\equiv3\pmod4\) be prime. Prove that the equation \(x^2+y^2=pz^2\) has only the zero solution in integers.
First prove \(p\mid x\) and \(p\mid y\), then obtain \(p\mid z\).
The equation implies \(p\mid x^2+y^2\). By the lemma for primes \(p\equiv3\pmod4\), we get \(p\mid x\) and \(p\mid y\). Then \(p^2\mid x^2+y^2=pz^2\), so \(p\mid z\). Dividing \(x,y,z\) by \(p\), we obtain a smaller integer solution. If a nonzero solution existed, iteration would give infinite descent. Therefore no nonzero solution exists, and the only solution is \(x=y=z=0\).
The closing problem shows how a modular obstruction becomes a full descent method.