If $m, n$ be positive integers, prove that $\phi(mn)=\phi((m,n))\phi([m,n])$, where $(m,n)=$ gcd of $m, n$ and $[m, n]=$ lcm of $m, n$. If $m, n$ be positive integers, prove that $\phi(mn)=\phi((m,n))\phi([m,n])$, where $(m,n)=$ gcd of $m, n$ and $[m, n]=$ lcm of $m, n$.


I have no idea to solve this question. Please help me to solve the problem.
It is false. Take $m=n$.

You can prove that the statement holds if and only if we are in the trivial case $\text{gcd}(m,n)=1$.

To be precise, the claim should be modified to
$$
\varphi(mn)=\mathrm{gcd}(m,n)\cdot \varphi(\mathrm{lcm}(m,n)).
$$



Proof: 

Since $mn=\mathrm{lcm}(m,n)\mathrm{gcd}(m,n)$ and a prime $p$ divides $mn$ if and only if it divides $\mathrm{lcm}(m,n)$ then
$$
\frac{\varphi(mn)}{mn}=\prod_{p\mid mn}\left(1-\frac{1}{p}\right)=\prod_{p\mid \mathrm{lcm}(m,n)}\left(1-\frac{1}{p}\right)=\frac{\varphi(\mathrm{lcm}(m,n))}{\mathrm{lcm}(m,n))}.
$$
It follows that
$$
\frac{\varphi(mn)}{mn}=\frac{\varphi(\mathrm{lcm}(m,n))}{mn} \cdot \mathrm{gcd}(m,n)
$$
hence the above claim..
First of all, we have $\operatorname{gcd}(a,b)\cdot\operatorname{lcm}(a,b)=ab$, then now you need to ascertain that Euler's Totient Function has the following property: $\phi(mn)=\phi(m)\phi(n)$. Hint: write a matrix with mn entries.