MathDB
Miklos Schweitzer 1962_4

Source:

September 18, 2008
floor functionmodular arithmeticquadraticssymmetryadvanced fieldsadvanced fields unsolved

Problem Statement

Show that \prod_{1\leq x < y \leq \frac{p\minus{}1}{2}} (x^2\plus{}y^2) \equiv (\minus{}1)^{\lfloor\frac{p\plus{}1}{8}\rfloor} \;(mod\;p\ ) for every prime p≡3  (<spanclass=′latex−bold′>mod</span>  4 ) p\equiv 3 \;(<span class='latex-bold'>mod</span>\;4\ ). [J. Suranyi]
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