I was working out some problems. This is giving me trouble.
- If $p$ is a prime number of the form $4n+1$ then how do i show that:
$ \sum\limits_{i=1}^{p-1} \Biggl( \biggl\lfloor{\frac{2i^{2}}{p}\biggr\rfloor}-2\biggl\lfloor{\frac{i^{2}}{p}\biggr\rfloor}\Biggr)= \frac{p-1}{2}$
Two things which i know are:
If $p$ is a prime of the form $4n+1$, then $x^{2} \equiv -1 \ (\text{mod} \ p)$ can be solved.
$\lfloor{2x\rfloor}-2\lfloor{x\rfloor}$ is either $0$ or $1$.
I think the second one will be of use, but i really can't see how i can apply it here.