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I know the following:

$$\varphi(2n)=\varphi(2)\varphi(n)=\varphi(n)\iff(2,n)=1$$ And $$\varphi(3n)=\varphi(3)\varphi(n)=2\varphi(n)\iff(3,n)=1$$

But now I'm not sure what to do with this info.

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Do you have any data about $\varphi$? E.g. is it an additve function? Continuous? Smooth? What are its domain and range? – Daniel Robert-Nicoud Oct 24 '13 at 17:18
write $n = 2^a 3^b m$ for $\gcd(m,6)=1,$ see what happens. – Will Jagy Oct 24 '13 at 17:19
Break up into cases: (i) neither $2$ nor $3$ divides $n$ (easy); (ii) both do; (iii), (iv) one does but not the other. – André Nicolas Oct 24 '13 at 17:22
@DanielRobert-Nicoud: I suspect that $\varphi$ is intended to be the Euler function. – Cameron Buie Oct 24 '13 at 17:29
All I was taught was that it's multiplicative. – agent154 Oct 24 '13 at 20:08

Consider the following: \begin{align*} \varphi(2n)&=\begin{cases} \varphi(2)\varphi(n)=\varphi(n)&\text{if $(2,n)=1$}\\ \varphi(2)\varphi(n)\frac{2}{\varphi(2)}=2\varphi(n)&\text{if $(2,n)=2$} \end{cases} \end{align*} and \begin{align*} \varphi(3n)&=\begin{cases} \varphi(3)\varphi(n)=2\varphi(n)&\text{if $(3,n)=1$}\\ \varphi(3)\varphi(n)\frac{3}{\varphi(3)}=3\varphi(n)&\text{if $(3,n)=3$} \end{cases} \end{align*} The equality $\varphi(2n)=\varphi(3n)$ only holds when $(2,n)=2$ and $(3,n)=1$. Therefore, $\varphi(2n)=\varphi(3n)$ for all $n$ that have a prime factor of $2$ and that do not have a prime factor of 3... Or $\forall n,\ s.t\ 2\mid n,\ 3\nmid n$.

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I think $\phi(3n)=3\phi(n)$ when $(3,n)=3$. – lhf Oct 25 '13 at 2:35
@lhf I think you may be right, but I don't know how to show it... My math above used the assumption that if $(3,n)=3$, then $\varphi(3n)=\varphi(3^2n)=\varphi(3^2)\varphi(n)=3(3-1)\varphi(n)$... but now I see that I didn't factor the $3$ out of the $n$, so I should probably have $\varphi(n/3)$ instead... – agent154 Oct 25 '13 at 3:08
See…. – lhf Oct 25 '13 at 3:11
@lhf Updated the math – agent154 Oct 25 '13 at 3:17

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