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 ). [J. Suranyi] floor functionmodular arithmeticquadraticssymmetryadvanced fieldsadvanced fields unsolved