More from California

I spent a pleasant hour in a cafe in Oakland reconstructing the elementary proof of the “supplementary law” (\frac{2}{p})=(-1)^{(p^2 -1)/8}.  This is exactly the kind of tricky elementary argument I have trouble remembering if I’ve only see it once before (which was the case).  Anyway, the proof goes as follows: let P = -1\cdot 2 \cdot -3 \cdot 4 \dots \pm\frac{p-1}{2}.  We’ll evaluate this mod p in two different ways.  On the one hand, we clearly have P=\prod_{i=1}^{(p-1)/2} (-1)^i i = \frac{p-1}{2}! \cdot (-1)^{\sum_{i=1}^{(p-1)/2} i} =\frac{p-1}{2}!\cdot (-1)^{(p^2 -1)/8}.  On the other hand, mod we have P = -1\cdot 2 \cdot -3 \cdot 4 \dots \pm\frac{p-1}{2} = (p-1) \cdot 2 \cdot (p-3) \cdot 4 \cdot \dots \cdot \frac{p \pm 1}{2} = 2 \cdot 4 \cdot 6 \cdot 8 \cdot \dots \cdot (p-3) \cdot (p-1) =\frac{p-1}{2}! \cdot 2^{(p-1)/2}.

Related puzzle: what is the value mod of f_p = 1\cdot 3 \cdot 5 \cdot \dots \cdot (p-4) \cdot (p-2)?  The first few values are misleading!

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