$ \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}$

Solution 1:

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

$$ \left \lfloor \frac{2i^2}{p} \right \rfloor - 2 \left \lfloor \frac{i^2}{p} \right \rfloor \quad \text{?}$$

Solution 2:

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!)