37th Iranian Mathematical Olympiad

The answer is $1$ for $m=0$, $p$ for $m=1$ and $\phi (p^m)$ for $m>1$.

Note that ${\mathbb{Z}}_p[x]$ is a unique factorization domain. So any monic $P(x)\in {\mathbb{Z}}_p[x]$ can be uniquely expressed as $P(x)\equiv A(x{)}^{2}Q(x)(\mathrm{mod}p)$, where $A(x),Q(x)$ are monic polynomials and $Q(x)$ is square-free. Let $f_n$ be the number of all square-free monic polynomials of ${\mathbb{Z}}_p[x]$ with degree $n$. Any non-square-free monic polynomial can be expressed as $P(x)=A(x{)}^{2}S(x)$ where $S(x)$ is a square-free monic polynomial.

If $\mathrm{deg}A(x)=m$, then $S(x)\in f_{n-2m}$. We have $f_{n-2m}$ options for $S(x)$ and $p^m$ options for $A(x)$. So $$f_n=p^n-\sum _{m=1}^{\lfloor \frac{n}{2}\rfloor }{f}_{n-2m}{p}^m.$$ And by simple induction on $n$ we get $f_n=\phi (p^n)$.