A quadratic equation over $\mathbb{Q}_p$

Question

Suppose we have the equation $x^2+x+1$ over the field $\mathbb{Q}_p$. is it possible to determine for what primes $p$ the equation has solutions?

I tried to see whether this is related to what $p$ is modulo $4$ by using quadratic reciprocity but it didn't work. I also looked for Hensel's lemma and didn't come up with anything. Do you have any suggestion for approaching this problem?

Answer 1

$(x-1)(x^2+x+1) = x^3-1$ so if $x^3 \equiv 1 \pmod p$, then $3|\phi(p) \implies p \equiv 1 \pmod 3$.

Note this holds for $p > 3$.

Answer 2

Two approaches come to mind.

• A quadratic has solutions (in a field of characteristic $\neq2$), iff its discriminant has a square root. The discriminant of your polynomial is $-3$, so....
• Also $x^3-1=(x-1)(x^2+x+1)$. So the zeros of your polynomial are primitive third roots of unity. For which primes does it happen that there is a primitive third root of unity in the prime field $\Bbb{Z}/p\Bbb{Z}$? What does Hensel tell you then?