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How can I prove for the Legendre symbol that:

$$\sum_{a=1}^{p-1}{\left(\frac{a(a+1)}{p}\right)}= -1 = \sum_{b=1}^{p-1}{\left(\frac{(1+b)}{p}\right)}$$

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Hint: Try adding and one to each sum, and reindexing the sum so that it looks like an orthogonality relation for characters.

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  • $\begingroup$ how will changing the bounds help? $\endgroup$ – kiki17 Mar 25 '13 at 23:31
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This is essentially what I prove here, after you note that $\left(\dfrac{(p-1)p}{p}\right)=\left(\dfrac{p}{p}\right)=0$. The main idea is to note that $\left(\dfrac{a(a+1)}{p}\right)=\left(\dfrac{\dfrac{a+1}{a}}{p}\right)$. Of course, you need to be familar with the fact that $\sum\limits_{a=1}^{p-1}{\left(\dfrac{a}{p}\right)}=0$ (which follows from the fact that there are exactly $\dfrac{p-1}{2}$ non-zero quadratic residues $\pmod{p}$ (for $p>2$))

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