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I know this question has been answered before, but I don't even know how to start this one.

I know $\phi$ is a multiplicative function, but I'm honestly so lost I don't know what to do.

Could anyone shed some guidance? Forgive my ignorance, please. I want to learn but I feel as though I've been left to fend for myself in this course

Martin Sleziak
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John L.
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    This has been answered here: http://math.stackexchange.com/questions/538378/solve-for-n-varphi2n-varphi3n Is there part of the answer you don't understand? – lulu Oct 08 '15 at 03:28
  • Do what they said to do: write $n=2^a3^bm$ with $2,3\nmid m$ and see what happens. – anon Oct 08 '15 at 03:31
  • I've looked at the answers and I just don't know how they decide to start where they do. :^(

    Whacka, I would try that if I knew how to make use of it.

    ϕ(2n) where n=2^a3^bm = (2^(a+1)-2^a)(3^b - 3^(b-1))(?) <- I don't know what to do with the m at all. if m is prime, then ϕ(m) = m-1, and if it's not? then what?

     – John L. Oct 08 '15 at 03:36
  • I know m cannot be even, as 2 does not divide it, but why would one even think to let n =(2^a)(3^b)(m)? – John L. Oct 08 '15 at 03:37
  • If $n = {2^a}{3^b}m$ with $\gcd(2,m) = \gcd(3,m) = 1$, then you have

    $$\phi(n) = \phi(2^a)\phi(3^b)\phi(m).$$

    It would suffice not to bother about $\phi(m)$, as the only information that you have at this point is that $\gcd(6,m) = 1$.

     – Jose Arnaldo Bebita Dris Oct 08 '15 at 03:41
  • Now, evaluate $\phi(2n)$ and $\phi(3n)$ separately (where it will be necessary to express them in terms of $\phi(m)$, by my previous comment). – Jose Arnaldo Bebita Dris Oct 08 '15 at 03:43
  • You don't do anything with the $ m$. Just compare $\phi (2 n)=\phi (2^{a+1})\phi (3^b)\phi (m)$ to $\phi (3 n)=\phi (2^a)\phi (3^{b+1})\phi (m)$ in the four cases $ (a=b=0),(a>0=b),(b>0=a),(a>0<b).$ – DanielWainfleet Oct 08 '15 at 03:43
  • So then, I must find a and b such that (2^(a+1) - 2^a)(3^b - 3^(b-1)) = (2^a - 2^(a-1))(3^(b-1) - 3^b)? – John L. Oct 08 '15 at 03:43
  • If $a=0$ we have $\phi (2 n)=\phi (n)$ but $\phi (3 n)= 2 \phi (n)$ when $b=0$, or $\phi (3 n)=3\phi (n)$ when $b>0$....... If $a>0<b$ we have $\phi (2 n)=2\phi (n)$ and $\phi (3 n)=3\phi (n)$......If $a>0=b$ then $\phi (2 n)=\phi (3 n)=2\phi (n).$ So it is necessary and sufficient that $n$ is even and not divisible by $3$. – DanielWainfleet Oct 08 '15 at 03:58
  • ...would you mind explaining why it is that you're able to make each conclusion that you do there? I have no idea why you conclude that if a=0 $\phi$($2n$) = $\phi$($n$) – John L. Oct 08 '15 at 04:02

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Hint: Euler's totient $\phi$ is an example of a (weakly) multiplicative function.

Jose Arnaldo Bebita Dris
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