# Families of quadratic polynomials over $\mathbb{F}_p$

Consider the set $$\mathcal{P}$$ of polynomials $$f \in \mathbb{F}_p[x, y]$$ in two variable over the field with $$p$$ elements, that are quadratic in $$x$$ and of unrestricted degree in $$y$$. For each fixed $$y \in \mathbb{F}_p$$ there are at most 2 solutions $$x$$ to the equation $$f(x, y) = 0$$. I will consider the subset of these polynomials, lets call it $$\mathcal{P}_{\textrm{nice}}$$, where this number is exactly 2 whenever $$y \neq 0$$:

$$\mathcal{P}_\textrm{nice} = \{ f \in \mathcal{P} : |\{x : f(x, y) = 0\}| = 2 \textrm{ for all } y \in \mathbb{F}_p^*\}$$

A really easy example of a polynomial in $$\mathcal{P}_\rm{nice}$$ is $$f(x, y) = x(x - y)$$ which for each non-zero $$y$$ has the two element zero-set $$x = 0$$, $$x = y$$.

A rather different example, for $$p = 5$$ is given by $$f(x, y) = x^2 + (y + 1)x - (y^2 + 2y + 2)$$ which for $$y = 0, 1, 2, 3, 4$$ gives the zero-sets $$\{1, 3\}, \{3, 0\}, \{0, 2\}, \{2, 4\}, \{4, 1\}$$. (I included the $$y = 0$$ case only to illustrate the nice cyclic structure on the zero-sets)

Generally speaking each element of $$f$$ gives us a list of pairs of elements of $$\mathbb{F}_p$$, indexed by elements of $$\mathbb{F}_p^*$$. The question is if every list of pairs, indexed by non-zero elements occurs in this way, so:

Let a list of $$p - 1$$ triples $$(a, b, c) \in \mathbb{F}_p^* \times \mathbb{F}_p \times \mathbb{F}_p$$ be given such that

1. $$a$$ runs through all elements of $$\mathbb{F}_p^*$$ and

2. $$b \neq c$$ for each fixed triple $$(a, b, c)$$ in the list.

Q: is it true that then there is always a polynomial $$f \in \mathcal{P}$$ such that for each $$(a, b, c)$$ in the list we have that $$f(b, a) = f(c, a) = 0$$?

(Note that such a polynomial would automatically lie in $$\mathcal{P}_\rm{nice}$$.)

It sounds to me like it should be false, but then I discovered (by brute force, when looking for something else, so rather accidentally) that it is true for $$p = 5$$.

Now I am a bit confused. If it is true, it sounds like there should be an easy way to construct the polynomial from the list of triples $$(a, b, c)$$, but I don't see it.

Given a set map $\theta:\mathbb{F}_p\to \mathbb{F}_p$ (or from $\mathbb{F}_p^*\to\mathbb{F}_p$), there exists a polynomial $f(y)\in\mathbb{F}_p[y]$ such that $\theta(a)=f(a)$ for all $a$. This is an elementary application of Van der Monde matrix.
Assuming this and given your $p-1$ triples, define $\theta_1(a)=-(b+c), \theta_2(a)=bc$. Then there exists polynomials $f(y), g(y)$ such that $\theta_1(a)=f(a), \theta_2(a)=g(a)$ for all $a$. Now consider the polynomial $x^2+f(y)x+g(y)$. Then for any $a$, we get the polynomial $x^2+f(a)x+g(a)=x^2-(b+c)x+bc=(x-b)(x-c)$, which is what you need.