Let p be a prime number with p = 1 mod 4. Prove that
sum(from a=1 to (p-1)/2) of (a over p) =0 where (a over p) denotes the legendre symbol.
We know that $\displaystyle \sum_{t=1}^{p-1}(t/p) = 0$.
Therefore, $\displaystyle \sum_{t=1}^{(p-1)/2} (t/p) + \sum_{t=(p+1)/2}^{p-1} (t/p) = 0$ .... [1]
However, $\displaystyle \sum_{t=(p+1)/2}^{p-1} (t/p) = \sum_{t=(p+1)/2}^{p-1} (-t/p) = \sum_{t=(p+1)/2}^{p-1} ((p-t)/p) = \sum_{t=1}^{(p-1)/2}(t/p)$
Therefore from [1] we get,
$\displaystyle 2\sum_{t=1}^{(p-1)/2} (t/p) = 0 \implies \sum_{t=1}^{(p-1)/2}(t/p) = 0$