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For $p\gt2$ a prime, and any $q \in \mathbb{Z^+} $,

Show that $\left(\frac{q(q+1)}{p}\right) =\left(\frac{1+q^{-1}}{p}\right )$ where the terms are legendre terms.

I saw this result as part of a proof, and it was simply assumed, like the result was trivial/obvious.. but I cannot seem to see why this is so.. is it a typo? or can someone please explain/prove why?

Thanks heaps!

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    Jake, if you want to be sure I see$a$comment addressed to me, you have to write @Gerry. Anyway, if, say, $p$ divides $a$, then multiplicativity still holds, as both sides are zero.2012-10-25

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