If $\phi(m) = \phi(mn)$ and $n \gt 1$, prove that $n = 2$ and $m$ is odd

Question

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

Answer 1

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.

Answer 2

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 $d = \gcd(m,n)$.