$ \sum\limits_{i=1}^{p-1} \Bigl( \Bigl\lfloor{\frac{2i^{2}}{p}\Bigr\rfloor}-2\Bigl\lfloor{\frac{i^{2}}{p}\Bigr\rfloor}\Bigr)= \frac{p-1}{2}$ 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.
 A: Here are some more detailed hints.
Consider the value of $\lfloor 2x \rfloor - 2 \lfloor x \rfloor$ where $x=n+ \delta$ for
$ n \in \mathbb{Z}$ and $0 \le \delta < 1/2.$
Suppose $p$ is a prime number of the form $4n+1$ and $a$ is a
quadratic residue modulo $p$ then why is $(p-a)$ also a quadratic residue?
What does this say about the number of quadratic residues $< p/2$ ?
All the quadratic residues are congruent to the numbers
$$1^2,2^2,\ldots, \left( \frac{p-1}{2} \right)^2,$$
which are themselves all incongruent to each other, so how many times does the set
$\lbrace 1^2,2^2,\ldots,(p-1)^2 \rbrace$ run through a complete set of
$\it{quadratic}$ residues?
Suppose $i^2 \equiv a \textrm{ mod } p$ where $i \in \lbrace 1,2,\ldots,p-1 \rbrace$ and $a$ is a quadratic residue $< p/2$ then what is the value of
$$ \left \lfloor \frac{2i^2}{p} \right \rfloor -
 2 \left \lfloor \frac{i^2}{p} \right \rfloor \quad \text{?}$$
A: Without giving everything away: when is $\lfloor2x\rfloor - 2\lfloor x\rfloor$ equal to $0$, and when is it equal to $1$?  Can you find some bijection between values of $i$ in your sum that fall into the first camp, and those that fall into the second?  (You may find the other fact you gave to be useful for finding this bijection!)
