If the same prime that divides $m$ and $n$, Prove that $n\varphi(m)=m \varphi(n)$
Then what is the relation between $m$ and $n$ Is it $m=p_1^{a_1}\cdots p_n^{a_n}\cdot 2^k$ And $n=p_1^{b_1}\cdots p_n^{b_n}$ ??
If yes what is the next step? any idea is highly appreciated
The result you're trying to prove can be rearranged as $\frac{\varphi(m)}{m}=\frac{\varphi(n)}{n}$, i.e. $\prod_{p\in\mathbb{P},\,p|m}(1-p^{-1})=\prod_{p\in\mathbb{P},\,p|n}(1-p^{-1})$. This is true if precisely the same primes divide $m,\,n$, but not necessarily just because they have one common prime factor. So we see that the $2^k$ factor you suggest is either wrong if none of the $p_i$ are $2$, or unnecessary if one of the $p_i$ is $2$.
Hint: If $n=p_1^{a_1}\dots p_k^{a_k}$ then $\phi(n)=n(\frac{p_1-1}{p_1})\dots(\frac{p_k-1}{p_k})$
Then $\frac{\phi(n)}{n}=(\frac{p_1-1}{p_1})\dots(\frac{p_k-1}{p_k})=\frac{\phi(m)}{m}$ as the same primes appear in the prime factorisation of $m$ and $n$
If $m$ and $n$ have the same prime factors, $p_1, ....... , p_k$ then
$m = \prod p_i^{a_i}$ for some set of positive integers $a_1, .... , a_k$ and $n = \prod p_i^{b_i}$ for some set of positive integers $b_1,...., b_k$.
The $\phi(m) = \prod (p_i - 1) \prod p_i^{a-1}$ and $\phi(n) = \prod (p_i - 1) \prod p_i^{b_i-1}$.
And $n\phi(m) = \prod p_i^{b_i}\prod (p_i - 1) \prod p_i^{a_i-1}=\prod (p_i - 1)\prod p_i^{a_i+b_i-1}$ and $m\phi(n) = \prod p_i^{a_i}\prod (p_i - 1) \prod p_i^{b_i-1}=\prod (p_i - 1)\prod p_i^{a_i+b_i-1}$.
So $m\phi(n) = n\phi(m)$
But notice that this may or may not be an if and only if statement.
You've only been asked to prove one direction.
