Feeds:
Posts
Comments

Archive for February 14th, 2008

O4. 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}.

Advertisements

Read Full Post »