Possible Duplicate:
Find the sum of all quadratic residues modulo $p$ where $p \equiv 1 \pmod{4}$
Show that if $p$ is a prime of the form $4k + 1$, the sum of quadratic residues ($\bmod p$) in the interval $[1, p)$ is $\frac{p (p - 1)}{4}$.
My attempt :
I could prove that the sum of the quadratic residues $\bmod p$ is $\frac{k ( 2k + 1 ) ( 4k + 1 )}{3} \bmod p$ (where $p = 4k + 1$). $p$ already divides this.
So, the sum of quadratic residues is a multiple of $p$.
What else can I try ?