
Mappings and Polynomials
Show that for every mapping $\displaystyle g:\mathbb{Z}/p\mathbb{Z}\rightarrow\mathbb{Z}/p\mathbb{Z}$, that there exists a polynomial $\displaystyle f(x)\in\mathbb{Z}/p\mathbb{Z}[x]$ such that $\displaystyle f(a) = g(a) $ for all $\displaystyle a\in\mathbb{Z}/p\mathbb{Z}$
I'm pretty sure Lagrange Interpolation is what is needed here but I'm not sure how to use it.

Yes since the sets used are finite with the same cardinal and that $\displaystyle \mathbb{Z}_p\{0\}$ is a group for $\displaystyle \times$ (I assume $\displaystyle p$ denotes a prime) you can write something like
$\displaystyle f:x\mapsto\sum\limits_{a\in \mathbb{Z}_p}\prod\limits_{k\in\mathbb{Z}_p\{a\}}\frac{(xk)g(a)}{(ak)}$ , and it is an element of $\displaystyle \mathbb{Z}_p[x]$