Let $f(x)=[1]x\in\mathbb{Z}_6[x]$. Find a polynomial $p(x)$ in $\mathbb{Z}_6[x]$ such that $p(x)\ne f(x)$, but $p\colon \mathbb{Z}_6\to \mathbb{Z}_6$ defines the same function as $f\colon \mathbb{Z}_6\to \mathbb{Z}_6$.
I am not sure what he is asking..? Expecially, $p\colon \mathbb{Z}_6\to \mathbb{Z}_6$ defines the same function as $f\colon \mathbb{Z}_6\to \mathbb{Z}_6$. this part makes me confused
Different polynomials in $\mathbb Z_m[X]$ can define the same function $\mathbb Z_m \to \mathbb Z_m$.
Fermat's theorem tells us that for $p$ a prime, the polynomial $x^{p}-x$ induces the zero function even though it is not the zero polynomial.
Hint: Try to find a non-zero polynomial $g$ such that $g(a)=0$ for all $a \in \mathbb Z_6$. Then take $p=f+g$.
Although $6$ is not a prime, look anyway for $g(x)=x^{n}-x$, for some $n$.
Solution:
$g(x)=x^3-x$ works. So does $g(x)=x^n-x$ with $n$ odd.
