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 nonzero element of $F$ is not a zerodivisor.

The same cannot be said of infinite fields. If $F$ is infinite, then there are functions $\varphi:F\to F$ that cannot be represented as polynomials. That's because the cardinality of $F[x]$ is the same as that of $F$ when $F$ is infinite. However, the number of functions $F\to F$ is greater than the cardinality of $F$. Therefore, there simply aren't enough polynomials.

But, one does not have to go to infinite fields. For any prime $q$, there are functions $\Z/q^2\to \Z/q^2$ that cannot be represented as a polynomial. This is true because if $\varphi$ is a polynomial function, then $\varphi(x + q)\equiv \varphi(x)$ modulo $q$. Therefore, any of the $q^{q-2}$ functions $\varphi:\Z/q^2\to\Z/q^2$ satisfying $\varphi(0) = 0$ and $\varphi(q) = 1$ cannot be represented by polynomial.