<s>[INST] A number with the form $n^2 +1$ , every of its odd divisor has the form $4k+1$ I dont know if it’s correct, so I ask you.
If we show, that every odd p which divides $n^2 +1$, has the form $4k+1$, then we are done.
So:
$$n^2\equiv -1\pmod{p} \implies (n^2)^{(p-1)/2}\equiv (-1)^{(p-1)/2}\pmod{p}\implies n^{p-1}\equiv 1\pmod{p}$$
If we substitute $p=4k+3$,
$$n^{4k+2}\equiv (n^2)^{2k+1}\equiv 1\pmod{4k+3}$$ which isn’t true, because $n^2\equiv -1\pmod{4k+3}$, and $2k+1$ is odd.
And if we subtitute $p=4k+1$ we get 
$$n^{4k}\equiv (n^2)^{2k}\equiv (-1)^{2k}\equiv 1\pmod{4k+1}$$ which is true.
 [/INST] If $p=4k+3$ then your first line is not correct because $\frac{p-1}{2}$ is odd.
Corrected proof
We can use Fermat's Little Theorem, $n^{p-1}\equiv 1\pmod{p}$. 
If we substitute $p=4k+3$, then
$n^{4k+2}\equiv (n^2)^{2k+1}\equiv 1\pmod{4k+3}$
which isn’t true, because $n^2\equiv -1\pmod{4k+3}$, and $2k+1$ is odd.
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