Problem
NT-B2-M03-P010 Order Three
#10
★★★☆☆ Level 3 of 5
Let \(p\) be prime, \(p\nmid a\), and \(p\mid a^2+a+1\). Prove that \(p=3\) or \(p\equiv1\pmod3\).
Multiply \(a^2+a+1\) by \(a-1\).
From \(a^2+a+1\equiv0\pmod p\), we get \(a^3-1=(a-1)(a^2+a+1)\equiv0\pmod p\). If \(a\equiv1\pmod p\), then \(3\equiv0\pmod p\), so \(p=3\). If \(a\not\equiv1\), the order of \(a\) is \(3\), hence \(3\mid p-1\), so \(p\equiv1\pmod3\).
This repeats the idea of \(a^2+ab+b^2\), but in one variable.