TITLE: Proof of a formula involving Euler's totient function: $\varphi (mn) = \varphi (m) \varphi (n) \cdot \frac{d}{\varphi (d)}$
QUESTION [26 upvotes]: The third formula on the wikipedia page for the Totient function states that $$\varphi (mn) = \varphi (m) \varphi (n) \cdot \dfrac{d}{\varphi (d)} $$
where $d = \gcd(m,n)$.
How is this claim justified?
Would we have to use the Chinese Remainder Theorem, as they suggest for proving that $\varphi$ is multiplicative?

REPLY [4 votes]: Just as an additional note to the identity provided by @Cardboard Box.
$$
\prod_{p|mn} (1 - \dfrac{1}{p}) = \frac{\prod_{p|m} (1 - \dfrac{1}{p}) \prod_{p|n} (1 - \dfrac{1}{p})}{\prod_{p|d} (1 - \dfrac{1}{p})}
$$
Why does this work?
Consider the prime factorization of $m,n$, and that we are multiplying all $p$ such that $p|mn$. But, on first sight, this is the same as multiplying all $p$ such that $p|n$ together with all $p$ such that $p|m$. But what if $m$ and $n$ share a prime factor?
This means that we will multiply that prime factor (call it $p_c$) twice, although of course it can only appear once in the prime factorization of $mn$, albeit with a greater power associated. 
Thus, we need to account for this shared factor, or even shared factors in case there is more than 1 shared prime factor. 
To do this, divide the whole thing by prime factors of the $gcd(m,n)$. Thus, we will be "removing" the factors which we multiplied twice. 
Furthermore, to go from 
$$
mn \frac{\prod_{p|m} (1 - \dfrac{1}{p}) \prod_{p|n} (1 - \dfrac{1}{p})}{\prod_{p|d} (1 - \dfrac{1}{p})} = \phi(m) \phi(n) \dfrac{d}{\phi(d)}
$$
consider multiplying the left side by $\dfrac{d}{d}$. Thus you can group together the $d$ with $\prod_{p|d} (1 - \dfrac{1}{p})$ in the denominator to get $\phi(d)$.
Hope this makes it easier to follow the top answer.