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I have to prove that the next equality holds: $$\sum_{k=0}^{p-1} \left( \frac{k(k+a)}{p} \right)=\sum_{k=0}^{p-1} \left( \frac{k(k+1)}{p} \right)$$ with $a \in \mathbb{Z}$ and $a$ not divisible by $p$ and p prime. I am supposed to use a substitution for this, but I have no idea which one.

Afterwards I have to use this equality, together with this one (which I already proved) $$\sum_{k,l=1}^{p-1} \left( \frac{kl}{p} \right)=0$$

to prove that $$\sum_{k=1}^{p-2} \left( \frac{k(k+1)}{p} \right)=-1.$$

Any help would be appreciated!

Bernard
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mathstudent
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2 Answers2

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Hint

If I understand you correctly, you cannot prove the first equality, nor use this to prove the second and thrid?

For the third note that for $1 \leq k \leq p-2$, $k$ has a unique inverse between $1$ and $p-2$. \begin{align} \sum_{k=1}^{p-2}{\left(\frac{k(k+1)}{p}\right)}&=\sum_{k=1}^{p-2}{\left(\frac{(\frac{k+1}{k})}{p}\right)}\\ &=\sum_{k=1}^{p-2}{\left(\frac{1+k^{-1}}{p}\right)}\\ &=\sum_{k=1}^{p-2}{\left(\frac{1+k}{p}\right)}\\ &=\sum_{k=2}^{p-1}{\left(\frac{k}{p}\right)}\\ &=-1 \end{align}

The first equality holds since \begin{align} \left(\frac{a(a+1)}{p}\right)&=\left(\frac{(\frac{a+1}{a})}{p}\right)\left(\frac{a^2}{p}\right)\\ &=\left(\frac{(\frac{a+1}{a})}{p}\right) \end{align} The last one also holds because $\sum\limits_{a=1}^{p-1}{\left(\frac{a}{p}\right)}=0 $ and $\left(\frac{1}{p}\right)=1$

  • Thanks for your answer, but I still have some questions. To prove the first equality I am supposed to use a substitution, I do not see how I can use your answer for that. Further I have to use the first and the second equality to prove the third. I understood your answer, but I am not supposed to do it that way since it does not use the first and second equality. – mathstudent Mar 13 '18 at 15:54
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$$\sum_{k\in\mathbb{F}_p}\left(\frac{k(k+a)}{p}\right)=\sum_{k\in a\mathbb{F}_p}\left(\frac{k(k+a)}{p}\right)=\sum_{k\in\mathbb{F}_p}\left(\frac{ak(ak+a)}{p}\right)=\sum_{k\in\mathbb{F}_p}\left(\frac{a^2}{p}\right)\left(\frac{k(k+1)}{p}\right)=\sum_{k\in\mathbb{F}_p}\left(\frac{k(k+1)}{p}\right).$$

Jack D'Aurizio
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