Quadratic residue $p \equiv 3 \pmod{4} $ 
Suppose $p$ is a prime congruent to $3$ modulo 4. Additionally, suppose $a$ is a quadratic residue modulo $p$. 
  Prove that $x=a^{\frac{p+1}4}$ is a solution to the congruence $x^2\equiv a \pmod  p$

I have no idea how to solve this. Can one use the Law of quadratic reciprocity ?
Since $a$ is a quadratic residue we know by eulers criterion that $a^{\frac{p-1}{2}} \equiv 1 \; mod  \; p$
I dont know how to use the condition that $p\equiv 3 \pmod{4}$ from here on

Let $x=a^{\frac{p+1}4}$. Then $x^2 =a^{\frac{p+1}2} = a^{\frac{p-1}2}\cdot a \equiv 1 \cdot a \bmod p$, by Euler's criterion.
The condition $ p \equiv 3 \bmod 4$ ensures that $\frac{p+1}4$ is an integer.
