Find a criterion for the odd primes p for which the congruence $x^2 + x + 3 \equiv 0 \pmod p$ has at least one solution. Explain how you know your answer is correct. A hint is given that the criterion can be expressed in values p modulo 11.
I know the most obvious path is to complete the square for quadratic reciprocity and use the legendre symbol, though I'm stuck on its actual execution, as well as actually checking how and why my answer is correct.
I did some guess work and ruled primes that divide the equation are $3,11$, and all primes where $p$ is congruent to $1,4,9,5,3\pmod {11}$ and primes that do not where $p$ is congruent to $2,6,7,8,10 \pmod {11}$. Am I at least kind of correct?
Thanks in advance homies.
First note that the congruence does not have a solution for $p=2$. For $p>2$ it has a solution if and only if $$(2x+1)^2+11\equiv0\pmod{p},$$ has a solution. Because $2$ is invertible mod $p$, this is equivalent to $-11$ being a quadratic residue mod $p$. By quadratic reciprocity you have $$\left(\frac{-11}{p}\right)=\left(\frac{-1}{p}\right)\cdot\left(\frac{11}{p}\right)=\bigg(\frac{p}{11}\bigg),$$ and so your congruence has a solution if and only if $p$ is a quadratic residue mod $11$, i.e. if and only if $$p\equiv0,1,3,4,5,9\pmod{11}.$$