Problem
NT-B2-M08-P013 Primes \(3 \pmod 4\)
#13
★★★★☆ Level 4 of 5
Let \(p\equiv 3\pmod 4\) be prime. Prove that if \(p\mid x^2+y^2\), then \(p\mid x\) and \(p\mid y\).
Assume \(p\nmid y\) and obtain a square root of \(-1\) modulo \(p\).
If \(p\nmid y\), then \((xy^{-1})^2\equiv -1\pmod p\). Raise both sides to \(\frac{p-1}{2}\). The left side is \(1\), while the right side is \((-1)^{(p-1)/2}=-1\), since \(p\equiv 3\pmod 4\). Contradiction. Hence \(p\mid y\), and then \(p\mid x\) follows from \(p\mid x^2+y^2\).
Connects to earlier modules: Fermat and quadratic residues.