Question about irreducible polynomials over finite fields

Question

I have the polynomial $f(T)=T^2+T+1$; then, for which primes $p$ does $f(T)$ have roots in $\Bbb F_p$?

I tried this way: since the three roots of $f(T)$ are generated from the cubic root of $1$, we need it to be contained in the field $\Bbb F_p$; namely that $m^3\equiv 1$ $($mod $p)$ for some $m\in \Bbb F_p$. For example in $\Bbb F_7$ we have that $2^3\equiv 1 $ $($mod $7)$ and in fact in this field $2$ is a root of the polynomial $f$; however I don't know how to describe in general in which fields $f(T)$ is reducible and in which is not.

Thank you :)

Answer 1

Hint: Use $4f(T)=(2T+1)^2+3$, and quadratic reciprocity.

Answer 2

You're just about there. As you've observed, $f(T)$ has roots in $\Bbb Z / p \Bbb Z$ if and only if $1$ has non-trivial cube roots in that field. And since the multiplicative group has order $\lvert (\Bbb Z / p \Bbb Z)^* \rvert = p-1$ and it's cyclic, that occurs exactly when $p \equiv 1 \pmod{6}$.