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Solving $\phi(m)$ =2 , to find that the only possible types of m are 3,4, and 6

I have considered the product and found $p_i$ = 2 or 3 and $a_i$ = 2 or 1 and this I have found using cases 1|2 and 2|2.

How can I use my results combining the two cases that the only possible such m are if 3,4, or 6.

edit: I am considering the case of when m and n are coprime and I know the possible solutions are (3,4) and (4,3) . For self, I am wondering about how I find these m after obtaining my cases 1|2 and 2|2.

R.ron
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1 Answers1

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If $p\mid m$ then $p-1\mid \phi(m)$, hence $p\le 3$.

If $p^k\mid m$, then $p^{k-1}\mid m$, hence $2^3\nmid m$ and $3^2\nmid m$.

This leaves us with $1,2,3,4,6,12$, which can be checked manually.

Hagen von Eitzen
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  • So in my case, I am unsure what you are suggesting. The only $p_i$ I have obtained are 2 and 3. I don't follow. – R.ron Oct 27 '17 at 16:59
  • I think I have roughly pieced it together. I did check cases manually.

    If you can elaborate more on what you were suggesting that would be cool to know.

     – R.ron Oct 27 '17 at 18:42