TITLE: Can all functions over $\mathbb{Z}/4\mathbb{Z}$ be "described" by a polynomial?
QUESTION [8 upvotes]: Consider the set $T$ of functions from $\mathbb{Z}/4\mathbb{Z}$ to $\mathbb{Z}/4\mathbb{Z}$. I'm now asked to prove or disprove the statement that all functions in $T$ can be described by a polynomial over $\mathbb{Z}/4\mathbb{Z}$, i.e. for any $f \in T$, there exists a polynomial $P \in (\mathbb{Z}/4\mathbb{Z})[X]$ so that the polynomial function for $P$ takes exactly the values that $f$ takes.
I already know that this statement is true if instead of $\mathbb{Z}/4\mathbb{Z}$, we have a finite field as the underlying structure, but since $\mathbb{Z}/4\mathbb{Z}$ is not a field, I rather suspect that this statement is not true, i.e. that there is any function that gets in our way. If I suspect that I have found such a function, I would need to show that no polynomial over $\mathbb{Z}/4\mathbb{Z}$ takes the same values as $f$.
I guess that there is a simple counterexample, but so far, I haven't been able to find one. Obviously we can't use any constant function, but I'm so far out of ideas what I can do to construct such a counterexample.

REPLY [3 votes]: Hint $\, $ If so there is a polynomial with $f(\color{#0a0}2)=0,\,f(0)=\color{#c00}1.\,$ Therefore, by the Factor Theorem, the root $\,\color{#0a0}2\,$ divides  the constant term $\color{#c00}1,\,$ i.e. $\,2a = 1,\,$ so $\,2\,$ is invertible, contra $\,2\cdot 2 = 0$.
Remark $\ $ Thus, if so, every nonzero element is invertible, i.e. the ring is a field.
Many nice proofs arise by exploiting relationships between the factorizations of polynomials and the factorization of the values they take. For example, one can devise a simple algorithm for polynomial factorization using factorization of its integer values and Lagrange interpolation. The ideas behind this algorithm are due in part to Bernoulli, Schubert, Kronecker.