Determine whether $x^2 \equiv 667\pmod{919}$ has solutions.
I'm not asking for an answer since I really want to figure this out, but researching and I can't find anything. Almost every example I find is that a is a perfect number such that $(x-a)(x+a)$, but in here The factors of $667$ are $1,23,29,667$ and $919$ is prime. Any links or hints to get me going?
$919$ is a prime $\equiv 3\pmod{4}$ and by the quadratic reciprocity theorem the Legendre symbol $\left(\frac{667}{919}\right)$ can be computed as follows: $$\left(\frac{667}{919}\right)=\left(\frac{23}{919}\right)\left(\frac{29}{919}\right)=-\left(\frac{-1}{23}\right)\left(\frac{20}{29}\right)=\left(\frac{5}{29}\right)=\left(\frac{-1}{5}\right)=+1$$ hence $x^2\equiv 667\pmod{919}$ has two solutions, namely $\pm242$.