show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$ $n = 4k + 3 $
We start by letting $a \not\equiv 0\pmod n$ $\Rightarrow$ $a \equiv k\pmod n$ .
$\Rightarrow$ $a^{4k+2} \equiv 1\pmod n$
Now, I know that the contradiction will arrive from the fact that if we can show $a^2 \equiv 1 \pmod n $ then we can say $b^2 \equiv -1 \pmod n $ then it is not possible since solution exists only for $n=4k_2+1 $ so $a \equiv b\equiv 0 \pmod n $
So from the fact that $a^{2^{2k+1}} \equiv 1 \pmod n$ I have to conclude something.

You are on the right track, the problem is that you don't know what $a^2$ is to relate it to $b^2$.
Hint Can you relate somehow $a^{4k+2} \equiv 1\pmod n$ to $b$?
Edit: Since you figured it, here is the complete solution. 
First note that if we prove $a \equiv 0\pmod p$ it follows imediately that $b$ is also $0 \pmod p$.
Thus it is enough to prove that $a \equiv 0 \pmod p$. Assume by contradiction that is not true. Then, since $p$ is prime, by Fermat Little Theorem, 
$$1=a^{p-1}=a^{4k+2}=(a^2)^{2k+1}=(-b^2)^{2k+1} =(-1)^{2k+1}b^{4k+2}=-b^{4k+2} (*)$$
But $b^2= -a^2$ is invertible, thus $b \neq 0 \pmod p$. Hence 
$$b^{4k+2}=b^{p-1} = 1 \pmod p$$
Thus $(*)$ implies $1=-1 \pmod p$, contradiction.
