Finding all polynomials $f(X), g(X)\in \mathbb{F}_5[X]$ such that $f(X)\neq g(X)$ and $f(x)=g(x)\neq 0$ for $x\in\mathbb{F_5}$.

Question

I have to find all polynomials $f(X),g(X)\in\mathbb{F}_5[X]$ such that $f(X)\neq g(X)$ and $f(x)=g(x)\neq 0$ for $x\in\mathbb{F}_5$ where $\mathbb{F}_5=\{0,1,2,3,4\}$.

For example: Let $f(X)=X^5$ and $g(X)=X$ where $f(X)\neq g(X)$ in $\mathbb{F}_5[X]$, but if I put $x\in\mathbb{F}_5$ in $f(x), g(x)$ then we get as follows:

• $x=0: f(0)=0^5 = 0 = g(0)$
• $x=1: f(1)=1^5=1 = g(1)$
• $x=2: f(2)= 2^5 = 32\equiv_5 2 = g(2)$
• $x=3: f(3)=3^5 \equiv_5 (-2)^5 = -32 \equiv_5 3 = g(3)$
• $x=4: f(4)=4^5 \equiv_5 (-1)^5 = -1 \equiv_5 4 = g(4)$.

So, for $x\in\mathbb{F}_5$, we get $x^5 =x$. How can I find other polynomials?

Answer 1

Just converting the comment to an answer :

Suppose $f(a)=g(a)$ , then $(f−g)(a)=0$ so $(x−a)$ divides $(f−g)$ as a polynomial. If $f(x)=g(x)$ for all $x \in \mathbb F_5$ then the above should tell you something about $f−g$ that turns out to be necessary and sufficient.

Basically, $f-g$ is a multiple of $\prod_{a \in \mathbb F_5} (x-a)$. Prove that this is sufficient and necessary.

Answer 2

If $f(x)=g(x)$ for all $x \in \mathbb F_5$, then $x, x-1,x-2,x-3, x-4$ divide $f-g$ and conversely.

So for example take any non vanishing polynomial for $f$ and $g(x) = f(x) + x^a (x-1)^b (x-2)^c (x-3)^d (x-4)^e$. This gives you an infinite number of examples.

Like

$$\begin{cases} f(x) &= x^2\\ g(x) &= x^2 + x(x-1)(x-2)^2(x-3)(x-4) \end{cases}$$