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04. PEN I11 (Korea 2000) Let $p$ be a prime number of the form $4k + 1$. Show that
$\sum^{p-1}_{i=1}\left( \left \lfloor \frac{2i^{2}}{p}\right\rfloor-2\left \lfloor \frac{i^{2}}{p}\right \rfloor \right) =\frac{p-1}{2}.$