For which primes $p\not=2$ is $5$ a square mod $p$? For which primes $p\not=2$ is $5$ a square mod $p$?

Using the Legendre symbol, $5$ is a square modulo $p$ if
$$\left(\frac{5}{p}\right)=5^{\dfrac{p-1}{2}} \equiv 1 \pmod{p}$$
Now we have
$$5^{\dfrac{p-1}{2}} = 
(5^{p-1})^{1/2} \equiv 1\pmod{p}$$
So $5$ is a square for any $p$. But this doesn't seem correct to me. Can anyone check this and if it's wrong, explain why.

You can't take the square root like this. By Fermat's little theorem, we know that for $p$ prime we have that $5^{p-1} \equiv 1 \mod p$. This means that $p \mid 5^{p-1} -1$, and hence $$p \mid \left(5^{\tfrac{p-1}{2}} -1\right)\left(5^{\tfrac{p-1}{2}} +1\right)$$
Therefore $5^{\tfrac{p-1}{2}} \equiv 1 \mod p$ or $5^{\tfrac{p-1}{2}} \equiv -1 \mod p$.
Note that if $p$ is not prime, there can be even more possibilities. 

Note that for every odd prime $p$ we have $2 \mid p-1$. Also $4 \mid 5-1$. So $2 \mid \left(\frac{p-1}{2} \cdot \frac{5-1}{2} \right)$. Hence by quadratic reciprocity we have $$\left( \frac{p}{5} \right)\left( \frac{5}{p} \right) = (-1)^{\frac{p-1}{2} \cdot \frac{5-1}{2}} = 1$$
So $\left( \frac{p}{5} \right) =1$ if $\left( \frac{5}{p} \right)=1$, and the latter happens when $p\equiv 1 \mod 5$ or $p \equiv 4 \mod 5$. 
Note that one can not use the law if $p=5$, since the law of  quadratic reciprocity requires $p$ and $q$ to be different primes. Whether $p=5$ is a quadratic residue depends on the definition of quadratic residue. 
