Let $p$ be an odd prime, and put $$s(a, p) = \sum_{n=1}^{p} \left(\frac{n(n+a)}{p}\right) $$
Show that:
(i) $s(0, p) = p − 1$
(ii) $\sum_{a=1}^{p}s(a, p) = 0.$
(ii) If $(a, p) = 1$ then $ s(a, p) = s(1, p).$
(iii) Conclude that $s(a, p) = −1$ if $(a, p) = 1.$
Is it enough for me for ii) to say that if $n>0$ and $s>0$ then the $\sum(s)>0$ hence cannot be true
Im not sure what to do for the other ones
Notice that $$ \sum_{n=1}^{p}\left(\frac{n(n+a)}{p}\right)=\sum_{n=1}^{p-1}\left(\frac{n^{-1}(n+a)}{p}\right)=\sum_{n=1}^{p-1 }\left(\frac{1+a n^{-1}}{p}\right)=\sum_{m=1}^{p-1}\left(\frac{1+am}{p}\right). $$ If $a$ is a multiple of $p$ the RHS is just $p-1$. Otherwise $m\mapsto am$ gives a permutation of $\mathbb{Z}/(p\mathbb{Z})^*$ and the RHS equals $$ \sum_{s=1}^{p-1}\left(\frac{1+s}{p}\right)=\sum_{s=1}^{p-2}\left(\frac{1+s}{p}\right)=-1+\sum_{s=1}^{p-1}\left(\frac{s}{p}\right)=-1. $$