I need to find all the integers such that $\phi (n) \mid n $, where $\phi$ is the totient function.

Using $$\phi(n)=n\prod(1-1/p)$$where the product runs over all prime factors of n, one gets that $$n/\phi(n)=\prod\frac{p}{p-1}$$ and this needs to be an integer. This is true if the list of primes is just {2} or {2,3}. I don't think there are other solutions, but I am having trouble showing it.



If $n$ has two (or more) odd prime factors, you will have at least two factors of $2$ in the denominator of your product, but they can't both be cancelled since the only possible even prime (in the numerator) is $2$.

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