Problem
ALG-B1-M09-P015 Prime divides a square
#15
★★★☆☆ Level 3 of 5
Let \(p\) be prime and \(p\mid a^2\). Prove that \(p\mid a\).
Consider the residue of \(a\) modulo \(p\).
If \(p\nmid a\), then \(a\) has an inverse modulo \(p\). From \(a^2\equiv0\pmod p\), multiplying by the inverse of \(a\) twice gives \(1\equiv0\pmod p\), impossible. Hence \(p\mid a\).
A basic fact for problems with squares and prime divisors.