Strangers in the night

The number of integers from $1$ to $n-1$ coprime to $n$ is the Euler's totient function $\phi(n).$ If $n=p_1^{k_1}\cdot ... \cdot p_s^{k_s}$ is the decomposition of $n$ into a product of powers of distinct primes, then $\phi(n)=(p_1-1)p_1^{k_1-1}\cdot ... \cdot (p_s-1)p_s^{k_s-1}=n\cdot (1-\frac{1}{p_1})\cdot ... \cdot (1-\frac{1}{p_s})$

Suppose $\phi(n)=132=11\cdot 12.$ Then $11\nmid n,$ otherwise $10$ would have to divide $132.$ Therefore, $11$ divides $p-1$ for some prime $p|n.$ Clearly, such $p$ must be of the form $22k+1,$ where $2k$ divides $12$ . Checking $k=1,2,3,4,6,$ we get $p=23,$ $p=67,$ $p=89$ ( $45$ and $133$ are not primes).

Case 1. $n=89\cdot m.$ Clearly, $89\nmid m,$ so $132=\phi(89) \cdot \phi(m),$ thus $\phi(m)=2.$ The only primes that could divide $m$ are $2$ and $3$ . Further, if $m=2^a3^b,$ then $a\leq 2$ and $b\leq 1.$ A simple check gives three possibilities: $m=3,4,6.$

Case 2. $n=67\cdot m.$ Clearly, $67\nmid m,$ so $132=\phi(67) \cdot \phi(m),$ thus $\phi(m)=3.$ This is impossible. In fact, no odd number greater than $1$ can be $\phi(m)$ for any $m$ : no odd prime can divide $m$ , and if $m=2^a,$ $\phi(m)=2^{a-1}.$

Case 3. $n=23\cdot m.$ Clearly, $23\nmid m,$ so $132=\phi(23) \cdot \phi(m),$ thus $\phi(m)=6.$ The only primes $p$ that can divide $m$ are $2,$ $3,$ and $7.$ If $7|m,$ then clearly $m=7$ or $m=14$ . Now suppose $m=2^a3^b$ . The order of $3$ in $\phi(m)$ is $b-1,$ so $b=2$ . Clearly, $m=9$ or $m=18.$

So altogether we have $7$ choices for $n$ .