How many numbers $m$ satisfy $1 ≤ m ≤ n$ and $\gcd (m, n) = 1$?

Question

Let $n = p^2 q$ where $p$ and $q$ are distinct prime numbers. How many numbers $m$ satisfy $1 \leq m \leq n$ and $\gcd (m, n) = 1$? Note that $\gcd (m, n)$ is the greatest common divisor of $m$ and $n$.

1. $p(q - 1)$
2. $pq$
3. $(p^2- 1) (q - 1)$
4. $p(p - 1) (q - 1)$

---

My attempt:

Using prime factorization :

Eulers function is $\phi(p^n)$ is $p^{n-1}(p-1)$

given that $n=p^2q(p,q$ prime$)$

$\phi(p^2q)=\phi(p^2)*\phi(q) =p(p-1)(q-1)$

Can you explain in alternative/formal way? Please.

Answer 1

as requested in the comments:

We can do this via the Principle of Inclusion/Exclusion.

There are exactly $pq$ numbers $1≤m≤p^2q$ which are divisible by $p$.

There are exactly $p^2$ which are divisible by $q$.

There are exactly $p$ which are divisible by both $p$ and $q$.

Thus: $$\varphi(p^2q)=p^2q-pq-p^2+p=p(p-1)(q-1)$$