Problem
NT-B2-M02-P018 Primes \(2\pmod3\)
#18
★★★★★ Level 5 of 5
Let \(p\equiv2\pmod3\) be prime and \(p\mid a^2+ab+b^2\). Prove that \(p\mid a\) and \(p\mid b\).
If one of \(a,b\) is not divisible by \(p\), reduce to the previous problem.
Assume not both numbers are divisible by \(p\). If \(p\mid b\), then the form gives \(p\mid a^2\), hence \(p\mid a\), contradiction. Thus we may assume \(p\nmid b\). Then \(t=ab^{-1}\) satisfies \(t^2+t+1\equiv0\pmod p\), and as in the previous problem this gives an element of order \(3\). Hence \(3\mid p-1\), so \(p\equiv1\pmod3\), contradicting \(p\equiv2\pmod3\). Therefore \(p\mid a\) and \(p\mid b\).
This is the analogue of the prime \(3\pmod4\) lemma for the form \(a^2+ab+b^2\).