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I want to show that

If $p$ is an odd prime, then $\sum_{ j = 1 }^{ p-2 } \left(\frac{j^2 + j}p \right) = -1$ (where $\left(\frac\cdot\cdot\right)$ is Legendre Symbol).

My teacher said I should use inverse of $j$ to solve this problem.

How can I solve this problem?

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We first prove the following result, which may be the trick using inverses hinted at by your teacher. For no good reason, we use $a$ instead of $j$. And we use the notation $L(x,q)$ for the Legendre symbol $\left(\frac{x}{q}\right)$.

Lemma: Let $p$ be an odd prime, and let $a$ be relatively prime to $p$. Let $b$ be the inverse of $a$ modulo $p$. Then $L(a(a+1), p)=L(1+b,p)$.

Proof: Since $1\equiv ab\pmod{p}$ we have $a(a+1)\equiv a(a+ab)=a^2(1+b)\pmod{p}$. But $L(a^2(1+b),p)=L(a^2,p)L(1+b,p)$. Since $L(a^2,p)=1$, the result follows.


Now as $a$ ranges from $1$ to $p-2$, the inverse $b$ of $a$ ranges, modulo $p$, over the numbers from $1$ to $p-1$. So $1+b$ ranges, modulo $p$, over the numbers from $2$ to $p-1$.

It follows that $\sum_{a=1}^{p-2} L(a(a+1),p)=\sum_{k=2}^{p-1} L(k,p)$.

But since there are just as many quadratic residues as non-residues, we have $\sum_{k=1}^{p-1} L(k,p)=0$. Since $L(1,p)=1$, we conclude that $\sum_{k=2}^{p-1}L(k,p)=-1$, and the desired result follows.

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