# Mappings and Polynomials

Show that for every mapping $g:\mathbb{Z}/p\mathbb{Z}\rightarrow\mathbb{Z}/p\mathbb{Z}$, that there exists a polynomial $f(x)\in\mathbb{Z}/p\mathbb{Z}[x]$ such that $f(a) = g(a)$ for all $a\in\mathbb{Z}/p\mathbb{Z}$
Yes since the sets used are finite with the same cardinal and that $\mathbb{Z}_p-\{0\}$ is a group for $\times$ (I assume $p$ denotes a prime) you can write something like
$f:x\mapsto\sum\limits_{a\in \mathbb{Z}_p}\prod\limits_{k\in\mathbb{Z}_p-\{a\}}\frac{(x-k)g(a)}{(a-k)}$ , and it is an element of $\mathbb{Z}_p[x]$