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$$\left(\frac{1\cdot2}{p}\right) + \left(\frac{2\cdot3}{p}\right) + \left(\frac{3\cdot4}{p}\right) +\ldots+ \left(\frac{(p-2)(p-1)}{p}\right) = -1$$

Note: $\left(\frac{a}{b}\right)$ represents the Legendre Symbol.

I have tried using this method. For each $k$ between $1$ and $p-2$ denote by $k'$ its multiplicative inverse $\mod p$. To estimate the sum of all Legendre symbols $\left(\frac{k(k+1)}p\right)$ show first that $\left(\frac{k(k+1)}p\right) = \left(\frac{1+k'}p\right),$ then estimate the sum of $\left(\frac{1+k'}p\right).$

Carlos Eugenio Thompson Pinzón
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user105693
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  • The last half has to be symmetric with the first half, so you get $-1=\left(\frac{(p^2-1)/4}{p}\right)+2\sum_{k=1}^{\frac{p-3}{2}} \left(\frac{k(k+1)}{p}\right)$, and since 1/4 is a QR (mod $p$) -1/4 must be a QNR if $p\equiv 3\pmod{4}$ and the problem reduces to $$ \sum_{k=1}^{\frac{p-3}{2}} \left(\frac{k(k+1)}{p}\right)=0$$ Or, if $p\equiv 1\pmod{4}$, $$ \sum_{k=1}^{\frac{p-3}{2}} \left(\frac{k(k+1)}{p}\right)=-1$$ – Tim Ratigan Nov 05 '13 at 20:40

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Your method is effective. We want to compute $$ \sum_{k=1}^{p-2} \left(\frac{k\cdot(k+1)}{p}\right)=\left(\frac{k^2}{p}\right)\left(\frac{1+k^{-1}}{p}\right)\sum_{k=1}^{p-2}\left(\frac{1+k^{-1}}{p}\right) $$ Because each nonzero residue $\pmod{p}$ has a unique inverse, and the inverse of $-1$ is $-1$, we obtain $$ \sum_{k=1}^{p-2}\left(\frac{1+k^{-1}}{p}\right)=\sum_{k=1}^{p-2}\left(\frac{k+1}{p}\right)=-1+\sum_{k=1}^{p-1}\left(\frac{k}{p}\right)=-1 $$ since $\left(\frac{1}{p}\right)=1$, and there are $\frac{p-1}{2}$ quadratic residues and non-quadratic residues each in the range $1,2,\cdots p-1$.

pi37
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