Find all primes $p$ such that $x^2\equiv 10\pmod p$ has a solution 
Find all primes $p$ such that $x^2\equiv 10$ mod $p$ has a solution

So I want $\left(\frac{10}{p}\right)=\left(\frac{5}{p}\right)\left(\frac{2}{p}\right)=1$
Then $\left(\frac{5}{p}\right)=\left(\frac{2}{p}\right)=1$
or $\left(\frac{5}{p}\right)=\left(\frac{2}{p}\right)=-1$
for the first The quadratic residues of $5$ are $1,4$
And $2$ is a quadratic residue for primes of the form $8k+1,8k+7$
so the primes of the form $5k+1,5k+4$ and either $8k+1,8k+7$ will work
or primes which are the form $5k+3,5k+2$ and $8k+3,8k+5$
I'm not sure if it's possible to narrow this down more?

Everything you do is correct (modulo noting that $\left(\frac{5}{p}\right) = \left(\frac{p}{5}\right)$ by Quadratic Reciprocity, which is why you are checking for quadratic residues modulo $5$) and dealing with the primes $p=2$ and $p=5$.
If $p=2$ or $p=5$, then $x^2\equiv 10\pmod{p}$ clearly has solutions.
Otherwise, your development is correct. You need either


*

*$p\equiv 1,-1\pmod{5}$ and $p\equiv 1,-1\pmod{8}$; hence $p\equiv \pm 1\pmod{40}$ or $p\equiv \pm 31\equiv\pm 9\pmod{40}$;

*$p\equiv 2,-2\pmod{5}$ and $p\equiv 3,-3\pmod{8}$; hence $p\equiv \pm 27\equiv \pm13\pmod{40}$ or $p\equiv\pm37\equiv \pm3\pmod{40}$;

*$p=2$;

*$p=5$.
