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.


1 Answer 1


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.

  • $\begingroup$ Wow, these Vandermondematrices are incredible! Thank you for pointing them out to me! $\endgroup$
    – Vincent
    Jun 7, 2018 at 14:36

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