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Archive for February 21st, 2008

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}.

Here is the official solution file: PEN04S
You can also disscuss the problems here!

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