Image of a polynomial function $x^2+y^2-x+y-axy$ over $\mathbb{F}_p$

Question

Let $p$ be an odd prime and $h(x)=x^2+ax+1$ be an irreducible polynomial over the field $\mathbb{F}_p$. I need to prove that the function

$$\Psi: \mathbb{F}_p^2 \longrightarrow \mathbb{F}_p, \quad (x,y)\mapsto x^2+y^2-x+y-axy$$

is surjective. I know that is true because the values in the image are in one-to-one correspondence with some conjugacy classes in some groups, but I would like to have an elementary proof of this fact, using the properties of polynomials over finite fields.

I tried to restrict to some suitable subset of the plane, like lines but I can not prove that the values the function takes when restricted to different lines cover all $\mathbb{F}_p$.

Answer 1

It does not matter whether $h$ is reducible or not, it is only important that $a\ne \pm 2$. By affine change of variables (start with $x-\frac a2 y=z$ and use new variables $(z,y)$, then shift them appropriately) we reduce the polynomial $\Psi$ to $x^2-by^2$, where $b=\frac{a^2}4-1$ is non-zero. This is surjective, since for any $\alpha\in \mathbb{Z}_p$ the sets $\{x^2,x\in \mathbb{Z}_p\}$ and $\{by^2+\alpha,y\in \mathbb{Z}_p\}$ have a common element: both contain more than a half of elements of $\mathbb{Z}_p$.