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I need help finding all $n$ such that $\phi(n)=n/3$.

I understand how to do it for $n/2$ but not for $n/3$. Thank you.

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  • $\begingroup$ Welcome to math.SE! This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. $\endgroup$ – user37238 Sep 18 '15 at 13:21
  • $\begingroup$ So you don't know how to do it for $n/2$, as your comment on the answer says. You 'think', but you're not sure. $\endgroup$ – user236182 Sep 18 '15 at 13:26
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Write $n=3^am$ with $3\nmid m$. Now $$\phi(n)=2\cdot 3^{a-1}\phi(m)$$ so you get $$\phi(m)=\frac{m}{2},$$ which you say you know how to solve.

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  • $\begingroup$ I think m = $2^k$ for some integer $k$. $\endgroup$ – asdfdsas Sep 18 '15 at 13:20
  • $\begingroup$ Let $m=2^km_1$ with $m_1$ odd. Then $\phi(m)=m/2=2^{k-1}\phi(m_1)$, so $\phi(m_1)=m_1$, which is only true if $m_1=1$, because $\phi(m_1)<m_1$ for $m_1\ge 2$. $\endgroup$ – user236182 Sep 18 '15 at 13:23
  • $\begingroup$ @ChrisRuscito Yes, but $m=2^k$ for some non-negative integer $k$. $\endgroup$ – user236182 Sep 18 '15 at 13:24

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