Find all the prime numbers $p$ which have a solution to $x^2\equiv14\bmod p$.

I found that for $p=2,5,7$ there is a solution and for $p=3$ there's no solution.
I tried using Legendre symbol because the given equation has a solution iff $(\frac{14}{p})=1 \iff (\frac{2}{p})(\frac{7}{p})=1$ but I get too many sub cases.


The "second supplement to quadratic reciprocity" states that $\left(\frac2p\right)=1\iff p\equiv\pm1\bmod8$. It remains to work out $\left(\frac7p\right)$. $$\left(\frac7p\right)\left(\frac p7\right)=(-1)^{3(p-1)/2}$$ Since $\left(\frac p7\right)=1\iff p\equiv 1,2,4\bmod7$ and the RHS is $1$ iff $p\equiv1\bmod4$, $\left(\frac7p\right)=1$ iff an even number of $p\equiv 1,2,4\bmod7$ and $p\equiv1\bmod4$ are true. Then $\left(\frac{14}p\right)=1$ iff an odd number of $p\equiv 1,2,4\bmod7$, $p\equiv1\bmod4$, $p\equiv\pm1\bmod8$ are true.

The last two conditions can be combined rather easily: an even number of them are true iff $p\equiv1,3\bmod8$. Thus $\left(\frac{14}p\right)=1$ iff an even number of $p\equiv1,2,4\bmod7$ and $p\equiv1,3\bmod8$ are true. By the CRT, this combines into $$p\equiv1,5,9,11,13,25,31,35,43,45,47,51\bmod56$$

All along we have ignored the exceptional cases, but the question already provides them. So we have the final result: $$\left(\frac{14}p\right)=1\iff p\equiv1,2,5,7,9,11,13,25,31,35,43,45,47,51\bmod56$$


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