For what integers $n$ does $\phi(2n)=\phi(3n)$?

Question

For what integers $n$ does $\phi(2n)=\phi(3n)$?

I know that $\phi(n) = \phi(P_1^{a1})\cdots\phi(P_k^{ak}) = (P_1^{a1}-P_1^{a1-1})\cdots(P_k^{ak}-P_k^{ak-1})$ but I'm not really sure how to apply it in a way that helps.

The only thing I could think to try was all cases of $n$ being even or odd, and $n$ being divisible by $3$ or not, however I could not get this to work.

Any help is appreciated, thanks.

Answer 1

Based on your formula for $\varphi(n)$, for any prime $p$,

$$\varphi(pn)= \left\{\begin{array}{ccc} (p-1)\varphi(n)& \text{if} & p\not\mid n \\ p\varphi(n) & \text{if} & p\mid n\end{array} \right. $$

If you apply this for $p=2$ and $p=3$ you should be able to come to something useful.

Answer 2

Hint: write $$ n = 2^a 3^b c $$with $2\nmid c, 3\nmid c$ and appy the formula.

Answer 3

If $3$ divides $n$, then $\varphi(3n)=3\varphi(n)\ne \varphi(2n)$.

If $3$ does not divide $n$, then $\varphi(3n)=2\varphi(n)$. This is $\varphi(2n)$ if and only if $2$ divides $n$.

These observations should be enough for you to finish.

Answer 4

Only for $n$ satisfying $n \equiv 2 \pmod 6$ or $n \equiv 4 \pmod 6$.

Remember that the totient function is multiplicative. This means that if $\gcd(a, b) = 1$, then $\phi(ab) = \phi(a) \phi(b)$. Also remember that $\phi(2) = 1$ and $\phi(3) = 2$, and the formula for $\phi(p^\alpha)$ where $p$ is some prime.

Let's look at this modulo 6.

• If $n \equiv 1 \pmod 6$, then $\phi(2n) = \phi(n)$ and $\phi(3n) = 2 \phi(n)$, so $\phi(2n) \neq \phi(3n)$.
• If $n \equiv 2 \pmod 6$, then $n = 2^\alpha m$ (where $\alpha$ is some integer and $m$ is odd), $\phi(2n) = 2^\alpha \phi(m)$ and $\phi(3n) = 2 \times 2^{\alpha - 1} \phi(m) = 2^\alpha \phi(m)$, so $\phi(2n) = \phi(3n)$ as desired.
• If $n \equiv 3 \pmod 6$, then $n = 3^\alpha m$ (where $\alpha$ is some integer and $m$ is not a multiple of 3), so we have $\phi(2n) = 2 \times 3^{\alpha - 1}\phi(m)$ and $\phi(3n) = 2 \times 3^\alpha \phi(m)$ (notice the exponents), so $\phi(2n) \neq \phi(3n)$.
• If $n \equiv 4 \pmod 6$, then $n = 2^\alpha m$ (where $\alpha$ is some integer and $m$ is odd), so we have $\phi(2n) = 2^{\alpha + 1} \phi(m)$ (because the 4 contributes 1 to 2's exponent) and $\phi(3n) = 2 \times 2^\alpha \phi(m) = 2^{\alpha + 1} \phi(m)$, so $\phi(2n) = \phi(3n)$ as desired.
• If $n \equiv 5 \pmod 6$, then $\phi(2n) = \phi(n)$ and $\phi(3n) = 2 \phi(n)$, so $\phi(2n) \neq \phi(3n)$.

Clearly odd numbers don't work for this. It should be easy enough for you to see that $n \equiv 0 \pmod 6$ does not work. That leaves us even numbers that are not multiples of 3.