$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.
This is solution from Prasolov V.V. Zadachi po algebre, arifmetike i analizu (В. В. Прасолов. Задачи по алгебре, арифметике и анализу.) This book (in Russian) is freely available at http://www.mccme.ru/free-books/
The problem appears there as Problem 31.2.
We want to show that if $p=4k+3$ is a prime number and $p\mid a^2+b^2$, then $p\mid a$ and $p\mid b$.
My translation of the solution given there:
Suppose that one of the numbers $a$, $b$ is not divisible by $p$. Then the other one is not divisible by $p$, either. Thus from Fermat's little theorem we get $a^{p-1} \equiv 1 \pmod p$ and $b^{p-1} \equiv 1 \pmod p$. This implies $a^{p-1}+b^{p-1} \equiv 2 \pmod p$.
On the other hand, the number $a^{p-1}+b^{p-1}=a^{4k+2}+b^{4k+2}=(a^2)^{2k+1}+(b^2)^{2k+1}$ is divisible by $a^2+b^2$, and thus it is divisible by $p$.
Hint: If $p \equiv 3 \pmod{4}$, then $x$ is a quadratic residue $\bmod{p}$ if and only if $-x$ is a quadratic nonresidue $\bmod{p}$.
Edit, to be more direct:
Lemma: Suppose that $p$ is an odd prime, and assume that $x$ is a nonzero quadratic residue $\bmod{p}$. Then, $-x$ is a quadratic residue $\bmod{p}$ if and only if $p \equiv 1 \pmod{4}$.
Proof:
First suppose that $p \equiv 1 \pmod{4}$. By assumption, as $x$ is a nonzero quadratic residue, there exists an integer $a \neq 0$ such that $a^2 = x$. Since $-1$ is also a quadratic residue, there also exists an integer $b$ such that $b^2 = -1$. Thus, $$ -x = -1 x = b^2 a^2 = (ba)^2, $$ and hence $-x$ is a quadratic residue.
On the other hand, suppose that $p \equiv 3 \pmod{4}$ and $x = a^2$ is a quadratic residue. Since $x \neq 0$, it follows that $-x \neq 0$ as well. Then, suppose for a contradiction that $-x$ is also a nonzero quadratic residue, and hence $-x = c^2$ for some $c$. If this were the case, it would follow that $$ -1 = -x/x = c^2/a^2 = (c/a)^2 $$ which would imply that $-1$ is a square $\bmod{p}$. This gives a contradiction, and we conclude that if $p \equiv 3\pmod{4}$ and $x$ is a nonzero quadratic residue, then $-x$ must be a quadratic nonresidue.
Assume that $a$ and $b$ are coprime. If $a$ and $b$ are odd, replace $a$ and $ b$ by $(a-b)/2$ and $(a+b)/2$. Then $a^2 + b^2 = pm$, and by reducing $a$ and $b$ modulo $p$ you can make sure that $m < p$. Since $p = 4n+3$ and $a^2 + b^2 = 4k+1$, we must have $m = 4j+3$. But then $m$ is divisible by a prime number $q = 4r+3$ strictly smaller than $p$. Repeat until you find that $3$ is a sum of two squares.
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.
