## All set endomorphisms of a finite field are polynomial

Let $F$ be a finite field. Did you know that given any function $\varphi:F\to F$, there exists a polynomial $p\in F[x]$ such that $\varphi(a) = p(a)$ for all $a\in F$? It's not hard to produce such the required polynomial: $$ p(x) = \sum_{a\in F} \left( \varphi(a)\prod_{b\not= a}(x – b)\prod_{b\not=a}(a-b)^{-1} \prod \right)$$ This works because every […]