Harvard-MIT Math Tournament

Solution:

$\varphi^{2001}\left(2002^{2}-1\right)=\varphi^{2001}(2001 \cdot 2003)=$ the number of $m$ that are relatively prime to both $2001$ and $2003$, where $m \leq 2003$. Since $\varphi(n)=n-1$ implies that $n$ is prime, we must only check for those $m$ relatively prime to $2001$, except for $2002$, which is relatively prime to $2002^{2}-1$. So $\varphi^{2001}\left(2002^{2}-1\right)=\varphi(2001)+1=\varphi(3 \cdot 23 \cdot 29)+1=(3-1)(23-1)(29-1)+1=1233$.