MathDB
I 11

Source:

May 25, 2007
floor functionquadraticsfunctionpen

Problem Statement

Let pp be a prime number of the form 4k+14k+1. Show that ∑i=1p−1(⌊2i2p⌋−2⌊i2p⌋)=p−12.\sum^{p-1}_{i=1}\left( \left \lfloor \frac{2i^{2}}{p}\right \rfloor-2\left \lfloor \frac{i^{2}}{p}\right \rfloor \right) = \frac{p-1}{2}.
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