Show $p$ prime s.t. $p \not\equiv 1 \mod 3$ is represented by the binary quadratic equation.

Question

I am working on the following question:

Let $p>3$ be a prime such that $p \not\equiv 1 \mod 3$. Show that $p$ is not represented by the binary quadratic equation $f(x, y) = x^2 + xy + y^2$.

I would appreciate any help.

Answer 1

If $p > 3$ is prime and not $1 \bmod 3$, it must be $2 \bmod 3$.

On the other hand, $f(x, y) = (x - y)^2 + 3xy \equiv 0, 1 \pmod 3$ since a square must be $0$ or $1 \bmod 3$.

Answer 2

Remember that you can just add up mods.

If $x \equiv 1 \bmod 3$ and $y \equiv 1 \bmod 3$, then $x^2 + xy + y^2 \equiv 0 \bmod 3$.

If $x \equiv 1 \bmod 3$ and $y \equiv 2 \bmod 3$, then $x^2 + xy + y^2 \equiv 1 \bmod 3$.

If $x \equiv 1 \bmod 3$ and $y \equiv 0 \bmod 3$, then $x^2 + xy + y^2 \equiv 1 \bmod 3$.

If $x \equiv 2 \bmod 3$ and $y \equiv 2 \bmod 3$, then $x^2 + xy + y^2 \equiv 0 \bmod 3$.

Obviously we needn't worry about both $x$ and $y$ satisfying $0 \mod 3$, and as for the other possibilities with $x$ and $y$ switched, we can ignore them "without any loss of generality."

Since $x^2 + xy + y^2 \equiv 2 \bmod 3$ is impossible, no $p \equiv 2 \bmod 3$ can be represented by it.