Casual Summation!

Note that $\left \lfloor \dfrac{2 k^2}{p} \right \rfloor - 2 \left \lfloor \dfrac{k^2}{p} \right \rfloor = \left \lfloor \dfrac{2 (k^2 \pmod p)}{p} \right \rfloor$ .

Now we have to find the number of solutions of $k^2 \pmod p > \left \lfloor \frac{p}{2} \right \rfloor$ .

Let $S_1$ be the set of integers $k$ from $1$ to $p-1$ which satisfy the condition, and $S_2$ be its complimentary set.

Now, $p \equiv 1 \pmod 4$ , so there is a solution to $x^2 \equiv -1 \pmod p$ . Let the solution be $a$ .

Note that $f(x) = a x \pmod p$ is an injection from $S_1$ to $S_2$ , and also from $S_2$ to $S_1$ .

Thus, $|S_1| = |S_2| = \dfrac{p-1}{2}$ , giving us the answer as $50.$

Where (.) denotes fraction part function and [.] denotes GIF.

Write $k^{2}/p = [ k^{2} /p] + (k^{2}/p)$ .

Multiply both sides by 2 and take GIF on both sides.

We write general term of this summation as

$\ [2(k^{2}/p)]$ .

Where (.) denotes fraction part function and [.] denotes GIF.

Now by induction we see that the summation of this quantity for $p=4\xi + 1$ is equal to

$[p/2]$ .

Note: Mathematical induction is applied on $\xi$ .