I was considering putting $\phi(mn)$ and $\phi(m)$ into their canonical form, but I'm not exactly too sure how this would help.
How would one approach this problem?
Thanks
I was considering putting $\phi(mn)$ and $\phi(m)$ into their canonical form, but I'm not exactly too sure how this would help.
How would one approach this problem?
Thanks
If you've already proven this result, you can just apply it and your result drops straight out: if $n > 2$, then $\phi(n) \geq 2$, and since $\gcd(m,n) \geq \phi(\gcd(m,n))$, we have $\phi(mn) \geq 2\phi(m) > \phi(m)$. Similarly, if $n = 2$ and $m is even$, then $\gcd(m,n) = 2$, so $\frac{\phi(m,n)}{\phi(m)} = 2\phi(n) \geq 2 > 1$.
If you haven't proven that result (and don't feel like just proving it to do this problem, we can proceed as you suggest:
$$\frac{\phi(mn)}{\phi(m)} = \frac{\displaystyle mn\prod\limits_{p|mn}\left(1 - \frac{1}{p}\right)}{\displaystyle m \prod\limits_{p|m}\left(1 - \frac{1}{p}\right)} = n\prod\limits_{\substack p|\,n\\p\not\,|\,m}\left(1 - \frac{1}{p}\right) \geq n\prod\limits_{p|n}\left(1 - \frac{1}{p}\right) = \phi(n).$$
Thus, if $\phi(mn) = \phi(m)$, then $\phi(n) = 1$, so $n = 2$.
But then we have $$\frac{\phi(2m)}{2\phi(m)} = \prod\limits_{\substack p|\,2\\p\not\,|\,m}\left(1 - \frac{1}{p}\right).$$
If $m$ is even, then that product is empty, so $\phi(2m) = 2\phi(m) \neq \phi(m)$ (since $m \neq 1$, since $1$ is not even). Thus, $m$ must be odd.
As an added bonus, we see that this condition is both necessary and sufficient: if $m$ is odd, then $\phi(2m) = \phi(2)\phi(m) = \phi(m)$, since $\phi$ is multiplicative.
Hint: This may be overkill, but it is well known that Euler's totient satisfies $$ \phi(mn) = \phi(m) \phi(n) \frac{d}{\phi(d)}, $$ where $d = \gcd(m,n)$.