Problem statement (and solution):
Let p be a prime satisfying $p \equiv 3 \pmod{4}$. Show that if the equation $x^2\equiv a \pmod{p}$ is soluble, then its solution is given by (1): $$x \equiv \pm a^{\frac{p+1}{4}} \pmod{p}$$
Solution: $x^2 \equiv a \pmod{p}$ implies that $(a\mid p) = 1$ So by Euler's criterion: $a^{\frac{p-1}{2}} \equiv 1 \pmod{p}$
if (1) holds, we have: $x^2 \equiv a^{\frac{p+1}{2}} \equiv a^{\frac{p+1}{2}} * a^{\frac{p-1}{2}} \equiv a^p \pmod{p}$
By Fermat's Little Theorem, $a^p \equiv a \pmod{p}$
This quite obviously also works in the other direction. My only real question is, why does the problem statement limit the space to primes congruent to 3 mod 4? Why would this exact proof not work for primes congruent 1 mod 4? What am I missing?