Problem source

For any positive integer $N$, the function $\f(N)$ is defined by
\[
\f(N) = N\Big(1-\frac1{p_1}\Big)\Big(1-\frac1{p_2}\Big)
\cdots\Big(1-\frac1{p_k}\Big)
\]
where $p_1$, $p_2$, $\dots$ , $p_k$ are the only prime numbers that are factors of $N$. 
Thus $\f(80)=80(1-\frac12)(1-\frac15)\,$.
\begin{questionparts}
\item
\textbf{(a)} Evaluate $\f(12)$  and $\f(180)$. 
\textbf{(b)} Show that $\f(N)$ is an integer for all $N$.
\item Prove, or disprove by means of a counterexample, each of the following:
\textbf{(a)} $\f(m) \f(n) = \f(mn)\,$;
\textbf{(b)} $\f(p) \f(q) = \f(pq)$ if $p$ and $q$ are distinct prime numbers;
\textbf{(c)}  $\f(p) \f(q) = \f(pq)$ only if $p$ and $q$ are distinct prime numbers.
\item Find a positive integer $m$ and a prime number $p$ such that $\f(p^m) = 146410\,$.
\end{questionparts}

Solution source

\begin{questionparts}
\item $f(12) = f(2^2 \cdot  3) = 12 \cdot (1-\frac12)\cdot(1-\frac13) = 12 \cdot \frac12 \cdot \frac 23 = 4$
$f(180) = f(2^2 \cdot 3^2 \cdot 5) = 180 \cdot \frac12 \cdot \frac23 \cdot \frac 45 = 12 \cdot 4 = 48$

Suppose $N$ has prime decomposition $p_1^{a_1} \cdots p_k^{a_k}$, then $f(N) = p_1^{a_1} \cdots p_k^{a_k} (1 - \frac{1}{p_1})\cdots ( 1- \frac{1}{p_k}) = p_1^{a_1-1} \cdots p_k^{a_k-1}(p_1-1) \cdots (p_k-1)$ which is clearly an integer.

\item $f(2) = 1, f(4) = 2 \neq f(2) \cdot f(2)$

If $p, q$ are distinct primes then $f(p) = p \cdot \frac{p-1}{p} = p-1$ and $f(q) = q-1$. $f(pq) = pq \frac{p-1}{p} \cdot \frac{q-1}{q} = (p-1)(q-1) = f(p)f(q)$
$f(12) = 4 = 2\cdot 2 = f(4) \cdot f(3)$

\item $f(p^m) = p^{m-1} (p-1)$
$146410 = 10 \cdot 14641 = 10 \cdot 11^4$. Therefore $p = 11, m = 5$
\end{questionparts}