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

There is no need to use quadratic reciprocity, and I do not think it is helpful even. 
Just try to check what is said: 
Is it true that $(a^{\frac{p+1}{4}})^2 = a \pmod{p}$?
That is, is it true that $a^{\frac{p+1}{2}} = a \pmod{p}$.
You have $a = b^2 \pmod{p}$ for some $b$ by assumption. So you'd need to show 
$b^{p+1} = b^2 \pmod{p}$. 
To do this, recall Fermat's Little Theorem.
