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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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$\left(\dfrac {q^2} p\right) = 1$ for all $q \in \Bbb Z^+$. Care needs to be taken that $q \ne 0 \pmod p$ for the RHS to make sense. – Lord_Farin Oct 25 '12 at 10:58
$q(q+1)=q^2(1+q^{-1})$. $(ab|p)=(a|p)(b|p)$. – Gerry Myerson Oct 25 '12 at 12:09
if we had GCD(q,p) = 1, does it necessarily follow that $GCD(1+q^{-1},p)=1$ also? since this would be required to use the rule: (ab|p)=(a|p)(b|p). – Jake Fresco Oct 25 '12 at 18:51
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. – Gerry Myerson Oct 25 '12 at 23:05

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