Suppose $Y(f)$ denotes the number of mutually incongruent solutions $(x,y)$ of the congruence $y^2 \equiv f(x) \pmod{p}$, where $p$ is an odd prime and $f(x)$ is a polynomial with integral coefficients. Prove that: $$ Y(f) = p + \sum_{n=0}^{p-1} \left( \frac{f(n)}{p} \right) $$
My thoughts:
We were given a hint that $Y(f)=\sum_{n=0}^{p-1}\left(1+\frac{f(n)}{p}\right)$ and I pulled out the one so that I had $Y(f)=1+\sum_{n=0}^{p-1}\frac{f(n)}{p}.$ I also know that $$ \sum_{n=0}^{p-1}\frac{(n-a)(n-b)}{p} = \begin{cases} p - 1 & \text{if }a \equiv b\ \pmod{p}\\ -1 & \text{otherwise} \end{cases} $$
I took $a$ to be $y^2$ and $b$ to be $f(x)$ thus concluding that the above sum will equal $p-1$. However, I am not sure if this is correct or if I am making the right approach. From here I do not know what else to do.
I guess, here $\left(\displaystyle\frac{f(n)}p\right)$ denotes the Legendre symbol, i.e. it is $0$ if $f(n)\equiv 0\pmod p$, it is $+1$ if $f(n)$ is any other quadratic residue, and $-1$ else.
I think, this explains it all, as if we add one, in all three cases above, we get that
Then, just run $x:=n$ from $0$ to $p-1$ and count the solutions (for $y$) of $\ y^2\equiv f(x) \pmod p$.