Let $F = GF(p)$, where $p$ is a prime integer, and let $g$ be an arbitrary function from $F$ to itself. Show that there exists a polynomial $f(X) ∈ F[X]$ of degree less than $p$ satisfying the condition that $g(c) =f (c)$ for all $c ∈ F$.
I think this problem means that every funtion $f(X)$ is different while $deg(f)<p$, so anyone can give an idea for how to prove that $f(c)=0\forall c\in GF(p),def(f)<p\Leftrightarrow f(X)=0$.
This is an exercise from the book: The Linear Algebra a Beginning Graduate Student Should Know by Golan.
See Lagrange interpolation. Alternatively you can check that the system of equations in the $p$ unknown coefficients of $f(X)$ has an invertible matrix. Hint: Its determinant is Vandermonde.