Let $p$ be an odd prime, and put $$s(a, p) = \sum_{n=1}^{p} \left(\frac{n(n+a)}{p}\right) $$
Prove that if $(a, p) = 1$ then $ s(a, p) = s(1, p).$
I have no idea how to prove this, I have tried using Wilsons Theorem but haven't really got anywhere, any help would be appreciated.
Altered version to avoid fractions, since they seem to cause confusion here.
Hint: Find $b$ so that $ab\equiv 1\pmod {p}$.
Then show that $$n(n+a)\equiv a^2\cdot nb(nb+1)\pmod{p}$$
So $$\left(\frac{n(n+a)}{p}\right)=\left(\frac{nb(nb+1)}{p}\right)$$