Suppose $p \equiv 1 \mod 3$. For $a \in \mathbb{Z}$ show that $f(x)=
x^3 – a$ has either $3$ or $0$ roots modulo $p$.
I've tried assuming you only have $1$ root. Then if $n$ is the root you can write $x^3 – a = (x-n)(x^2 + xn + n^2)$. Now I need a contradiction in assuming that quadratic has no root.
Best Answer
You are on the right track. Suppose there is at least one root, $n$. As you have noted, there is a factorization $x^3 - a = (x-n)(x^2 + xn + n^2)$. Now, to see that the quadratic has a root, you need to show that the discriminant $(-3n^2)$ is a square modulo $p$, or equivalently that $-3$ is a square modulo $p$. And remember, you need to use the mod $3$ condition. Does quadratic reciprocity ring a bell?