Problem
NT-B2-M02-P017 A Form of Order \(3\)
#17
★★★★★ Level 5 of 5
Let \(p\ne3\) be prime, \( \gcd(a,b)=1 \), and \(p\mid a^2+ab+b^2\). Prove that \(p\equiv1\pmod3\).
Divide the congruence by \(b^2\) and obtain \(t^2+t+1\equiv0\).
Since \(p\nmid b\), put \(t\equiv ab^{-1}\pmod p\). Then \(t^2+t+1\equiv0\). Multiplying by \(t-1\), we get \(t^3-1\equiv0\). Also \(t\not\equiv1\), otherwise \(3\equiv0\pmod p\), impossible. Hence the order of \(t\) is \(3\). The order divides \(p-1\), so \(3\mid p-1\), meaning \(p\equiv1\pmod3\).
This is advanced because the method does not look like ordinary quadratic residues.