# Casual Summation!

$\large S(p) = \sum_{k=1}^{p-1} \left( \left \lfloor \dfrac{2k^2}{p} \right \rfloor - 2 \left \lfloor \dfrac{k^2}{p} \right \rfloor \right)$

Let $$p$$ be a prime number such that $$p \equiv 1 \pmod4$$. Find the closed form of $$S(p)$$.

Enter your answer as $$S(101)$$ correct up to 2 decimal places.



Notation: $$\lfloor \cdot \rfloor$$ denotes the floor function.

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