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(I didn't find any other posts related to this but I think this is a weird because my question seems like a very natural one.)

Let $\varphi$ denote the Euler Totient Function. Given $m$ an even number, can we find all the positive integers $n$ such that $\varphi(n)=m$? If not, at least can we tell how many there are? Are there infinite? Why?

I strongly believe this is a hard task.

Thank you in advance.

Shoutre
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    You are looking at the inverse of the totient function, which happens to be multi-valued. Do have a look at this similar question on the number of $n$ which satisfies $\phi (n) = m$ for a given $m$: http://mathoverflow.net/questions/109509/the-inverse-of-the-euler-totient-function – Yiyuan Lee Oct 17 '15 at 01:50
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    There won't be infinitely many solutions. There can't be any prime factors of bigger than $m+1$, and for each other prime $p$, if $p^{k}>m+1$ then $p^{k+1}$ cannot be a factor of $n$. – Thomas Andrews Oct 17 '15 at 01:58

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Finite. $$ \varphi(n) \geq \sqrt {\frac{n}{2}} $$

More complete statements and a proof as two answers at Is the Euler phi function bounded below?

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Will Jagy
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