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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