# Let $\phi$ be Euler's totient function. Prove that $\phi(3n) = 3\phi(n)$ iff 3 divides n.

I managed to find that $\phi(3n) = 3\phi(n) \implies 3|n$ via contradiction:

Suppose that 3 does not divide n. Then gcd(3, n)=1, because 3 is a prime. Which, in turn, implies that $\phi(3n) = \phi(3)\phi(n)$ since Euler's totient function is multiplicative.

Also, note that $\phi(3) = 2$.

However, by hypothesis, $\phi(3n) = 3\phi(n)$, therefore we arrive at the conclusion that $3\phi(n) = 2\phi(n)$

$\phi(n)>0, \forall n \implies 3=2$. Contradiction!
$\therefore 3|n$

However, I can't find my way back. I have tried writing n as $n=3^\alpha m$, where 3 does not divide m. Therefore, $3n = 3^{\alpha+1}m$, which implies that $\phi(3n) = (3^{\alpha+1}-3^\alpha)\phi(m)=3^\alpha(3-1)\phi(m)$ but that only takes me as far as $\phi(3n) = 2.3^\alpha \phi(m)$, and I see no way of progressing (or if that even is the right path).

Does anyone have any suggestions? (Also, it's my first post here, any feedback is very welcome!)

• If $n=3^am$ as you have written (with $a≥1$) , then $\varphi(n)=3^{a-1}\times 2 \times \varphi (n)$, Now look at $\varphi(3n)$. – lulu Apr 8 '18 at 19:52
• Wouldn't that be $\phi(n)= 3^{\alpha-1}.2.\phi(m)$ ? The new $m$ in the totient function is what's throwing me off. – Pedro Orlando Apr 8 '18 at 20:13
• Yes, that's what I meant. – lulu Apr 8 '18 at 20:25

Write $n=3^\alpha m$ with $\alpha\ge0$ and $\gcd(3,m)=1$ and compute:
$\phi(3n) = \phi(3^{\alpha+1}m) = \phi(3^{\alpha+1})\phi(m)$
$3\phi(n) = 3\phi(3^{\alpha})\phi(m)$
Now note that $\phi(3^{\alpha+1}) = 3\phi(3^{\alpha})$ iff $\alpha>0$, that is, iff $3$ divides $n$.
The same argument works for any prime: $\phi(pn) = p\phi(n)$ iff $p$ divides $n$.