Weird quadratic residue question. Show that is p is a prime s.t $ p \equiv 3 \mod 8 $ and $\frac{(p-1)}{2}$ is also prime then show that $\frac{(p-1)}{2}$ is a quadratic residue of p. 
Now i am able to solve the problem except when $\frac{(p-1)}{2} \equiv 3$ i mean i believe that $\frac{(p-1)}{2} \equiv 1$ but i can't actually prove it.
 A: If $p$ is an odd prime, $\frac{p-1}{2}$ is a quadratic residue iff $-2$ is a quadratic residue.
Let us assume $p\equiv 3\pmod{8}$ and consider the splitting field of $\Phi_8(x)=x^4+1$ over $\mathbb{F}_p$.
Its degree over $\mathbb{F}_p$ is given by the least $k\in\mathbb{N}^+$ such that $8\mid(p^k-1)$, i.e. $2$.
It follows that $\Phi_8$ factors over $\mathbb{F}_p$ as the product of two quadratic irreducible polynomials.
Let us denote through $i$ and $\sqrt{2}$ the elements of $\mathbb{F}_p$ or $\mathbb{F}_{p^2}$ solving $x^2+1=0$ and $x^2-2=0$.
Let us consider the irreducible factor of $\Phi_8$ vanishing at $\frac{1+i}{\sqrt{2}}$. By Frobenius automorphism, the conjugated root is $\left(\frac{1+i}{\sqrt{2}}\right)^p = \frac{-1+i}{\sqrt{2}}$ where the last equality follows from $p\equiv 3\pmod{8}$.
Thus we have that one of the irreducible factors of $\Phi_8$ is 
$$ \left(x-\frac{-1+i}{\sqrt{2}}\right)\left(x-\frac{1+i}{\sqrt{2}}\right)=x^2-\sqrt{-2}\,x-1 $$
and since the coefficients of this polynomial belong to $\mathbb{F}_p$, $-2$ is a quadratic residue $\!\!\pmod{p}$.
$\frac{p-1}{2}$ being a prime is irrelevant.
A: i wasn't going to post this but the other solution is so complicated i felt compelled to post it. (despite the fact that it is actually a much stronger solution) 
My solution needs  $\frac{(p-1)}{2} $ to be prime to use the Legendre symbol's and the fact that $ 4= 2^2 $ is a obviously a quadratic residue.
$(\frac{(p-1)}{2}/p)=(2(p-1)/p) \equiv (\frac{-1}{p})(\frac{2}{p})  $ 
Since $p\equiv 3 \mod 8 $ we have $(\frac{2}{p}) =-1 $ by theorem. 
Since $p\equiv 3 \mod 8 \to p\equiv 3 \mod 4 \to (\frac{-1}{p})=-1 \to (\frac{-1}{p})(\frac{2}{p})=(-1)(-1)=1 $
$\therefore (\frac{(p-1)}{2}/p)=1 $ thus $ \frac{(p-1)}{2} $ is a quadratic residue of p as desired.
