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!
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