$\xi_p =e^{2\pi/p*i }$ , proof that is only two quadratic extention to $\Bbb Q $ that contained in $Q(\xi_p)$ such that $\Bbb Q(\sqrt p)$ if $p=1 mod (4)$ or
$\Bbb Q(\sqrt{-p})$ when $p=3mod(4)$
In the class we proof that $p*(-1/p)=z^2 $ when $z=\sum_{a\in \Bbb F^x_p}(a/p)*\xi_p^a$
when (-1/p) = 1 if p=1 mod 4 and
=-1 if p=3 mod 4
so it`s how I proof the statment. but how to show that this is the only two that possible according to the statment? I try by contradiction for $z\in \Bbb Q $ such that $\sqrt z \not= \sqrt p $ so $\sum_{i=1}^N a_i*\xi_p^i=\sqrt z$ for some $a_i\in \Bbb Q $ but Its lead me to nowhere...
