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On Jan. 18, 2012 I conjectured that for any prime $p>3$ we have $$R_p^2\equiv\frac1{10}\left(512\left(\frac{10}p\right)-27\left(\frac{-15}p\right)-475\right)\pmod p,$$ where $(\frac{\cdot}p)$ ...

Let $p>3$ be a prime, and let $(\frac{\cdot}p)$ be the Legendre symbol. Then $$14\left(\frac 3p\right)-\left(\frac p3\right)-12=\begin{cases}1&\text{if}\ p\equiv1\pmod{12}, \\-25&\text{if}\ ...