$q \equiv 2 (\mod 3)$ be a prime , then does there exist non-zero integers $a,b,c$ such that $a^2+ab+b^2=qc^2$ ?

Question

Let $q$ be a prime such that $q \equiv 2 (\mod 3)$ , then is it true that $a^2+ab+b^2=qc^2$ has no solution in non-zero integers $a,b,c$ ?

Answer 1

With the prime $q\equiv2\pmod3$ the only way $a^2+ab+b^2$ can be divisible by $q$ is for both $a$ and $b$ to be divisible by $q$. This is seen as follows. Assume contrariwise that $q\nmid b$. Then in the field $\Bbb{F}_q$ $b$ is invertible. Dividing the congruence $a^2+ab+b^2\equiv 0\pmod q$ by $b^2$ then implies that $x=a/b\in\Bbb{F}_q$ is a solution of the equation $$ x^2+x+1=0. $$ Clearly $x\neq1$, because $a=b$ is impossible. But $x^3-1=(x-1)(x^2+x+1)=0$ then implies that $x$ is of multiplicative order $3$.

But the multiplicative group of the field $\Bbb{F}_q$ is of order $q-1$ and, by Lagrange's theorem, cannot have an element of order three.

This settles your question as follows. Clearly $q\mid qc^2$, so $a^2+ab+b^2$ must be divisible by $q$. But then both $q\mid a$ and $q\mid b$ by what we just saw, so the L.H.S. is divisible by $q^2$. Therefore $q^2\mid qc^2$ meaning that $q\mid c$ also. But we can cancel the square of any common factor of $a,b,c$ from your equation. So assuming that this was done at the beginning we have arrrived at a contradiction.