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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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4… and – Beni Bogosel Apr 23 '12 at 12:24
Use the formula for the totient function in terms of the prime factors of n. – Dayo Adeyemi Apr 23 '12 at 12:32
@dayo Adeyemi: I do this and find all the prime factors must be consecutative, therefore can only be $2$ or $3$? – Freeman Apr 23 '12 at 12:35
@LHS think about it. Can $n$ be prime? or Can $n$ be odd? – Kirthi Raman Apr 23 '12 at 12:42
@Kv Raman: n can't be prime, unless it equals the totient function, however I'm assuming it can't be odd either? – Freeman Apr 23 '12 at 12:47
up vote 16 down vote accepted

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

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

If $n>2$, 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 $n=2^a3^b$, with $a>0$, $b\ge0$ is a solution.

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This is a very helpful answer, thanks so much! I understand this concept much better now – Freeman Apr 23 '12 at 13:29
I feel a bit bad about this. This was meant to address the point left open in m.k.'s answer. But while I was typing, that was deleted. I guess there is a lesson to be learned here? – Jyrki Lahtonen Apr 23 '12 at 13:35
Ah. Well I'm very grateful to m.k. As well. Hope they read this! – Freeman Apr 23 '12 at 14:03
@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. – Bill Dubuque Apr 23 '12 at 19:40

Quasi-brute-force approach using Maple :

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