suppose p is fermat prime and p>3, show that $\left(\frac{7}{p}\right) = -1$ suppose p is fermat prime, show that $\left(\frac{7}{p}\right) = -1$
We wish to solve $\left(\frac{7}{p}\right) = -1$ for $p$. By quadratic reciprocity and the fact that $7 \equiv 3 \pmod{4}$ we know $\left(\frac{7}{p}\right) = - \left(\frac{p}{7}\right)$. 
By a simple calculation, we see that $\left(\frac{7}{p}\right) = -1$ if $p \equiv 1,2,4 \pmod{7}$.
I know that fermat primes are equal to $p \equiv 1 \pmod{4}$ either. 
How can i use this information to conclude that $\left(\frac{7}{p}\right) = -1$?

Fermat primes are of the form $p=2^{2^{k}}+1$ for some $k\in\Bbb Z_{\ge 0}$. If $k=0$, your claim is wrong; but it's true if $k\ge 1$. I'll assume $k\ge 1$.
We know $p\equiv 1\pmod{4}$, so by Quadratic Reciprocity $\left(\frac{7}{p}\right)=\left(\frac{p}{7}\right)$.
We know $2^{2^k}+1\equiv \{5,3\}\pmod{7}$, none of which is a quadratic residue, so $\left(\frac{p}{7}\right)=-1$.
