Solve $\phi(n) = 123456789 $ or prove there is no such $n$

Question

I've solved it, but I'm interested to see if there is another way, because mine is computer assisted. Using the fact that if $p \mid n$ then $ p-1 \mid \phi(n) $, I find the set of divisors for 123456789: $$1,3, 9, 3607, 3803, 10821, 11409, 32463, 34227, 13717421, 41152263, 123456789$$ From this set only $1+1$ is prime, so $n = 2^a$. But $\phi( 2^a ) = 2^{a-1}$, but $123456789$ is not even, therefore such $n$ does not exist.

I used a script to find the divisors, I'm wondering if there is a not too complicated way to get the same result without "external" help. Thanks!

EDIT: Edited my proof, I forgot $\phi$ is not completely multiplicative.

score 17 · Accepted Answer · edited Apr 13 '17 at 12:21

17

$\phi(n)$ is even for $n\ge 3$, and obviously $\phi(1),\phi(2) \neq 123456789.$

edited Apr 13 '17 at 12:21

Community

1

answered Apr 20 '15 at 12:37

Casteels

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Well that was simple, that property is quite handy, thanks! – Xsy Apr 20 '15 at 12:43
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Indeed, $\phi(n)$ is even for $n>2$. – Gerry Myerson Apr 20 '15 at 13:04

Solve $\phi(n) = 123456789 $ or prove there is no such $n$

1 Answers1