# Question about irreducible polynomials over finite fields

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 :)

• note: $4$ is also a root in $\mathbb F_7$ – J. W. Tanner Jun 24 at 23:08
• The cube root is a slight distraction because $1$ is always a cube root of $1$. – user10354138 Jun 24 at 23:09
• $f$ has only at most two roots in any field… – Bernard Jun 24 at 23:13

Hint: Use $$4f(T)=(2T+1)^2+3$$, and quadratic reciprocity.
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 $$p \equiv 1 \pmod{6}$$.