Let 𝑝 be an odd prime and $ℎ(𝑥)=𝑥^2+𝑎𝑥+1$ be an irreducible polynomial over the field $\mathbb{Z}_p$. The polynomial function

$$\Psi:\mathbb{Z}_p^2⟶\mathbb{Z}_p,\quad (𝑥,𝑦)\mapsto 𝑥^2+𝑦^2−𝑥+𝑦−𝑎𝑥𝑦$$

is surjective, as proved here: Image of a polynomial function $x^2+y^2-x+y-axy$ over $\mathbb{F}_p$.

I would like to compute a set of representatives of the classes of the kernel of Ψ (i.e. the relation $𝑘𝑒𝑟(\Psi)={((𝑥,𝑦),(𝑡,𝑤))\in \mathbb{Z}_p^4, \, \Psi(𝑥,𝑦)=\Psi(𝑡,𝑤)})$. So basically I would like to provide a set of $𝑝−1$ elements of $\mathbb{ℤ}_p^2$ that have all different values when you applies $\Psi$ (I am not interested to the case $\Psi(𝑥,𝑦)=0$).

Is there a way to do that in general, regardless who is $p$?

Let 𝑝 be an odd prime and assume $𝑥^2+ax+1$ is irreducible over the field $\mathbb{F}_p$. The polynomial function

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

is surjective, as proved here: Image of a polynomial function $x^2+y^2-x+y-axy$ over $\mathbb{F}_p$.

I would like to compute a set of representatives of the classes of the kernel of $\Psi$ (i.e., the relation $\ker(\Psi)=\{(x,y,t,w)\in \mathbb{F}_p^4, \, \Psi(𝑥,𝑦)=\Psi(t,w)\})$. So basically I would like to have an explicit set of $p-1$ elements of $\mathbb{F}_p^2$ that take on all the nonzero values in $\mathbb{F}_p$ when you apply $\Psi$. (Finding a solution to $\Psi(x,y)=0$ is obvious.)

Is there a way to do this in general, regardless of the value of $p$?