Respected all.
The following problem I got stick although elementary but I am unable to finish it through logical argument.
Question is: If $m, n \in \mathbb N$ then $\sum\limits_{d|(m, 2n), d\neq 1, 2}\varphi(d)$ is same as $\sum\limits_{d|(m,n)}\varphi(d)$. How to establish it ? Is it true in general or there is any counter example for it ?
Any kind of help will be appreciated
As stated in the comment, we only need to ask wether $(m,2n)-2=(m,n)$ holds or not. Write $D:=(m,n)$. Since $D$ is a common divisor to $m$ and $2n$ we can find an integer $l$ such that :
$$(m,2n)=Dl $$
The equation becomes :
$$D(l-1)=2 $$
But $l$ cannot be anything else than $2$ or $1$, hence $l$ is necessarily $2$ if the above equation holds. This implies that $D=2$. On the other hand it is easily seen that if $D=2$ and $l=2$ then the above equation holds. I think that the set of solution is :
$$(2^{k}m',2n')\text{ where } m' \text{ and } n' \text{ are odd and prime to each other and } k\geq 2 $$
$\sum_{1\le d\mid m}\varphi(d)=m$ for any $m\in\Bbb Z^+$ (3 proofs here).
The equation is equivalent to $(m,2n)-2=(m,n)$.
Define $\upsilon_2(a)=t\iff (2^t\mid a,\, 2^{t+1}\nmid a)$.
$(m,2n)>(m,2n)-2=(m,n)$ and $2$ is prime,
so we must have $(m,2n)=2(m,n)$ and $\upsilon_2(m)>\upsilon_2(n)$, i.e.
$(m,2n)-2=(m,n)\implies ((m,n)=2,\, \upsilon_2(m)>\upsilon_2(n))$.
$\iff (m=4k,\, n=2n_1$ with $(k,n_1)=1$, $\,n_1$ odd$)$,
which is indeed a solution.
