Can all functions over $\mathbb{Z}/4\mathbb{Z}$ be "described" by a polynomial? 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.

Consider a polynomial $p(X) = \sum_{k=0}^n a_k X^k$ with $a_k \in \mathbb{Z}/4\mathbb{Z}$. Then we have
$$ p(0) = a_0, \\
   p(2) = a_0 + a_1 \cdot 2 + a_2 2^2 + \dots + a_n 2^n = a_0 + 2a_1. $$
Hence, we can't find a polynomial that satisfies for example $p(0) = 1, p(2) = 2$ as the equation $2a_1 = 1$ has no solution in $\mathbb{Z}/4\mathbb{Z}$.
