Let $\F_q$ be a finite field. For any function $f:\F_q\to \F_q$, there exists a polynomial $p\in \F_q[x]$ such that $f(a) = p(a)$ for all $a\in \F_q$. In other words, every function from a finite field to itself can be represented by a polynomial.

In particular, every permutation of $\F_q$ can be represented by a polynomial. This isn't true for other finite rings, a fact which is the topic of one of my papers. At any rate, polynomials that represent permutations are called **permutation polynomials**.

It's not easy to determine *which* polynomials are permutation polynomials. The exception is for the finite field $\F_2$, and more generally $\Z/2^w$. Then this case Ronald Rivest gave a straightforward criterion for whether a polynomial is a permutation polynomial.

However, there is a classical result that says any permutation may be represented *uniquely* by a polynomial whose degree is at most $q-2$. Here is an example noted by Charles Wells. Let $a,b\in\F_q$ be two distinct elements. The polynomial

$$f(x) = x + (a-b)(x-a)^{q-1} + (b-a)(x-b)^{q-1}$$

represents the transposition of $a$ and $b$. I suggest that you verify this by direct substitution, using the identity $c^{q-1} =1$ whenever $c\not=0$, which in turn follows from Fermat's little theorem.

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