Find all primes $p > 11$ such that $11$ is a quadratic residue $\pmod p$. (The answer should be in terms of congruence conditions mod a certain number.)
I believe that we need to find all primes $p > 11$ such that $ \left( \frac{11}{p} \right) = 1$.
or
$\left( \frac{11}{p} \right) = (-1)^{(p-1)/2}\left( \frac{p}{11} \right)$ by the reciprocity law.
How do I proceed from this? Any help is appreciated.
Best Answer
By quadratic reciprocity:
$$ \left(\frac{11}{p}\right) = (-1)^{(p-1)/2} \left(\frac{p}{11}\right)$$
We now need to work with two cases:
This means that:
$$ \left(\frac{11}{p}\right) = \left(\frac{p}{11}\right)$$
The quadratic residues modulo $11$ are $1, 3, 4, 5$ and $9$.
This means that:
$$ \left(\frac{11}{p}\right) = -\left(\frac{p}{11}\right)$$
The quadratic non-residues modulo $11$ are $2, 6, 7, 8$ and $10$.
With this information, you should be able to finish the problem yourself (hint: use the Chinese Remainder Theorem).