Quadratic residue p-1 modulo p for p primes 
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*It seems that for every prime p = (4k + 1), p-1 is a quadratic residue modulo p.

*It seems that for every prime p = (4k - 1), p-1 is not a quadratic residue modulo p.


I can't figure how to proove this, does anyone has an idea ? 
Best regards
 A: Your question is: For $ p \equiv 3 \mod 4$, $p-1$ is not a quadratic residue modulo $p$. 
Recall Euler's Criterion:
$a$ is a quadratic residue if and only if $a^{\frac{p-1}{2}} \equiv 1 \mod p$. 
 (I can supply a proof of this if necessary).
Now for the case of $p-1$ we see that $p-1 \equiv -1 \mod p$,  so our original question is equivalent to asking whether $-1^{\frac{p-1}{2}} \equiv 1 \mod p$. Notice that if $p \equiv 3 \mod 4$.
Then $p - 1 \equiv 2 \mod 4$, and $\frac{p-1}{2} \equiv 1 \mod 4$. In particular $\frac{p-1}{2}$ is odd. 
Thus $(-1)^{\frac{p-1}{2}} = -1$. Thus ${(p-1)}^{\frac{p-1}{2}} \equiv (-1)^{\frac{p-1}{2}} \equiv -1 \mod p$. Thus, by Euler's Criterion, $p-1$ is not a quadratic residue modulo $p$ if $p \equiv 3 \mod 4$.   
A similar argument shows that if $p \equiv 1 \mod 4$, $p-1$ is a quadratic residue modulo $p$. 
A: Every odd prime $p$ has a primitive root $g$ (actually many) such that the smallest $k>o$ for which $g^k \equiv 1 \bmod p$ is $k=p-1$. 
In that case, $g^{(p\mathord-1)/2} \equiv -1 \bmod p$. If $(p\mathord-1)/2$ is even then there is a value $a\equiv g^{(p\mathord-1)/4}$ such that $a^2\equiv -1 \bmod p$ and $-1$ is a quadratic residue $\bmod p$. If $(p\mathord-1)/2$ is odd, there is no such number and $-1$ is not a quadratic residue $\bmod p$. The parity of $(p\mathord-1)/2$ translates into a statement about whether $1$ or $3 \equiv p \bmod 4$.
