For positive integers $m$ and $n$, if $$\phi(mn)=\phi(m)$$ and $n>1$, then $n=2$ and $m$ is odd, where $\phi$ denotes the Euler totient function.
State with justification whether the statement is true or false. In my opinion I think the statement is true but I can't find the starting point for the proof.. it's really confusing.
On
Unique prime factorization is your friend.
Let $\gcd(m,n) = \prod r_i^{v_i}$ where $r_i$ are prime.
Let $m = \prod p_i^{w_i} \prod r_i^{u_i}$ are the $p_i$ are prime distinct from $r_i$ and $u_i \ge v_i$
Let $n = \prod q_i^{j_i} \prod r_i^{z_i}$. ditto. And $z_i > u_i$ only if $u_i = v_i$ and vice versa.
So $\phi(nm) = \prod q_i^{j_i-1}\prod (q_i-1)* \prod p_i^{w_i - 1}\prod (p_i-1) \prod r_i^{max(u_i,z_i)-1}\prod (r_i-1)$.
And $\phi(m) \prod p_i^{w_i - 1}\prod (p_i-1) \prod r_i^{\max(u_i,v_i) - v_i}\prod (r_i-1)$
Sooooo.... $\prod q_i^{j_i-1}*\prod (q_i - 1)*\prod r_i^{u_i - v_i} = 1$.
So all $q_i - 1 = 1$ so $q_i$ can only be $2$. And $\prod q_i^{j_i-1} = 2^{j_1-1} = 1$ so $j_1= 1$. So $m = 2$.
The $\gcd(2, m)$ must be either $2$ (if $m$ is even) or $1$ if $m$ is odd. So $r_i = 2$ and $u_i=v_i=1$ or $0$.
If $m = 2m'$ where $m'$ is odd then $\phi(nm) = \phi(2m) = \phi(2^2m') = \phi(2^2)\phi(m')$ whereas $\phi(m) = \phi(2m') = \phi(2)\phi(m')$. But this implies $\phi(2^2) = 2 = \phi(2) = 1$. So $m = m'$ is odd.
$\textbf{Hint}$:
$$\phi(mn)=\phi(m)\phi(n) \frac{d}{\phi(d)}$$
where $d=\text{gcd}\;(m,n)$
[For the proof, refer Apostol's Introduction to Analytic Number theory]
$\textbf{Addition}$:
By this result, and your hypothesis
$$\phi(m)=\phi(m) \phi(n) \frac{d}{\phi(d)}$$
So $\phi(n) \frac{d}{\phi(d)}$ must be $1$. Since both factors are positive integers, $\phi(n)=1$ and $\frac{d}{\phi(d)}=1.$
$\phi(n)=1$ implies $n=1$ or $2$, but $n>1$ so $n$ must be 2.
Also$$\frac{d}{\phi(d)}= \frac{gcd(m,n)}{\phi(gcd(m,n))}=\frac{gcd(m,2)}{\phi(gcd(m,2))}=1$$ implies $m$ must be odd
Hope this helps!