Find all primes $p$ such that $14$ is a quadratic residue modulo $p$. I want to find all primes $p$ for which $14$ is a quadratic residue modulo $p$. I referred to an example that was already posted for finding all odd primes $p$ for which $15$ is a quadratic residue modulo $p$, but I am getting stuck.
This is what I have done:
$$\left(\frac{14}{p}\right)=\left(\frac{7}{p}\right)\left(\frac{2}{p}\right)=(-1)^{(p-1)/2}\left(\frac{p}{7}\right)\left(\frac{p}{2}\right).
$$
There are now two cases. If $p\equiv 1\pmod{13}$, you have $(14|p)=(p|7)(p|2)$, want $(p|7)$ and $(p|2)$ to have the same sign. The squares modulo $2$ and $7$ are $p\equiv 1\pmod{2}$ and $p\equiv 1,2,4\pmod{7}$, and the nonsquares are $p\equiv 3,5,6 \pmod{7}$. 
This is where I am stuck do I just check the possibilities $\pmod {182}$? I think I need to use the condition that $p \equiv 1\pmod{13}$. I am just very confused and would really appreciate some clarification.

Actually quadratic reciprocity says
$$\color{Blue}{\left(\frac{7}{p}\right)}\color{Green}{\left(\frac{2}{p}\right)}=\color{Blue}{(-1)^{(p-1)/2}\left(\frac{p}{7}\right)}\color{Green}{(-1)^{(p^2-1)/8}}=(-1)^{(p^2+4p-5)/8}\left(\frac{p}{7}\right).$$
Therefore you want $A=(-1)^{(p^2+4p-5)/8}$ and $B=\left(\frac{p}{7}\right)$ to have the same sign. When is $A=+1$ and when is $A=-1$? When is $B=+1$ and when is $B=-1$? In all four cases, you should obtain congruence conditions. Then glue them together appropriately with the remainder theorem.
