Quadratic Reciprocity Law

Question

Can anyone help me find the solution to the quadratic equation $y^2 \equiv 1+8q^{p-1} (mod \ p)$ where p and q are primes such that p>q?

I did this by trial and error but does anyone know a systematic solution to solve this quadratic equation?

Answer 1

$$y^2\equiv1+8q^{p-1}\bmod p.$$ Now, $p$ is prime, so by Fermat's little theorem $q^{p-1}\equiv 1\bmod p$. So $$y^2\equiv 1+8=9\bmod p,$$ and so $$y\equiv \pm 3\bmod p.$$