This exercise popped up in a chapter on Legendre symbols.
Let $p$ be a prime $3$ $(\textrm{mod } 4)$, and $f(x) \in \mathbb{F}_p[x]$ a polynomial of odd degree. Show that the number of solutions $(x, y)$ in $\mathbb{F}^2_p$ of $y^2 = f(x)$ equals $p$.
I can understand some partial results to this problem, but I don't see a clear road into a general proof. Obviously every zero leads to a single solution. When considering the polynomial in the splitting field as something along the lines of $f(x) = a \cdot \prod^n_{i=1}(x - k_i)^{r_i}$ then all "surplus" even powers can be ignored and the problem reduces to finding an $x$ such that the leftover $a \cdot (x - k_{i_1}) \dots (x - k_{i_m})$ is a square. For example in the very specific case of a polynomial $f(x) = a \cdot (x - k_1)^2 (x - k_2)$ you get the solutions $(k_1, 0), (k_2, 0)$, and I can see for any $z \in \mathbb{F}_p$ that $x = k_2 + a^3 \cdot z^2$ leads to yet another perfect square, some of which ofcourse are duplicates (or $x = k_2 + z^2$ if $a$ itself is a quadratic residue). However I cannot figure out how this all comes together in the general case. I also don't quite understand how Legendre symbols might facilitate the exercise, I'm guessing maybe that's my biggest issue.
Also whenever you find a solution $(x, y)$, then $(x, -y)$ is a solution as well. Since $p$ is odd, wouldn't this imply that the polynomial $f(x)$ needs to have at least one zero?
Edit: I have managed to get a hold of the author. For completion's sake I will post the correct exercise below and follow it up with a solution.
Let $p$ be a prime $3$ $(\textrm{mod } 4)$, and $f(x) \in \mathbb{F}_p[x]$ a polynomial that is odd (i.e. $f(-x) = -f(x)$). Show that the number of solutions $(x, y)$ in $\mathbb{F}^2_p$ of $y^2 = f(x)$ equals $p$.
Since $f(x)$ is odd, $f(0) = -f(0) = 0^2$. Now let $x_1, \dots, x_k, x_{k+1}, \dots, x_{p-1}$ denote all the non-zero elements of $\mathbb{F}_p$, and assume $x_1, \dots, x_k$ are all zeroes of the polynomial $f(x)$. Then obviously these zeroes are part of the solution as $f(x_i) = 0^2$. Because $f(x)$ is odd, $f(-x_i) = -f(x_i) = 0^2$. This means $x_i \in x_1, \dots, x_k$ if and only if $-x_i \in x_1, \dots, x_k$, and therefore $k$ must be even. Since $p \equiv 3 \textrm{ mod } 4$, $-1$ is not a quadratic residue. Consequently, for every $j > k$, it follows that $(\frac{f(x_j)}{p}) = (\frac{-f(-x_j)}{p}) = -(\frac{f(-x_j)}{p})$, and $x_{k+1}, \dots, x_{p-1}$ contains exactly as many residues as non-residues. However for every of these $x_j \in x_{k+1}, \dots, x_{p-1}$, if $f(x_j) = y^2$ then also $f(x_j) = (-y)^2$; each element contributes either zero or two solutions. Consequently, the amount of solutions is given by $1 + k + 2 \cdot \frac{p - 1 - k}{2} = p$.