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Where $\phi(n)$ is the Euler phi function, how do you find all $n$ such that $\phi(n)|n$?

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    @LHS That is correct, I just wanted to see if you could think further.2012-04-26

2 Answers 2

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Notice that $\varphi(1) = \varphi(2) = 1$, so $\varphi(1) \mid 1$ and $\varphi(2) \mid 2$.

If $n > 2$, assume that the prime factorization of $n$ is

$n = p_1^{a_1} \ldots p_k^{a_k}$

Then the formula for the totient function gives

$\varphi(n) = (p_1 - 1)p_1^{a_1-1}\ldots (p_k - 1)p_k^{a_k-1}.$

Since $n>2$, this is always an even number, so $p_1=2$ must appear as a factor. We next observe that $n$ cannot have two odd prime factors. If $a_2>0$ and $a_3>0$, then both $p_2-1$ and $p_3-1$ are even, so $2^{a_1+1}\mid \varphi(n)$, which is a contradiction.

So $n=2^{a_1}p^{a_2}$ for some prime $p>2$. Here $p-1\mid\varphi(n)\mid n$, so $p-1$ must be a power of two, say $p-1=2^\ell$. Then $2^{a_1-1+\ell}\mid\varphi(n)$, so we must have $\ell=1$ and $p=3$.

In the end we can verify that $n=1$ or $n=2^a3^b$, with $a>0$, $b\ge0$.

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    @LHS There are at least a couple prior questions on this topic, so you might find some other prior answers also of interest. Please link them into this question if you find them.2012-04-23
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Quasi-brute-force approach using Maple :

with(numtheory): for n from 1 to 100 do if n mod phi(n) = 0 then print(n); end if; end do;