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I need to prove for a given n, if $\phi(x)=n$ has a solution for x, it always has another?

We know $\phi(2)=\phi(1)=1$ and can easily prove that n must be even for x>2.

So, n can be of the form $2^a.q$ where a>0 , odd q are natural numbers.

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I see no question mark anywhere... – J. M. Aug 12 '12 at 11:53
@lab bhattacharjee I need to contact with you regarding some discussion. Can you please mail me your contact id to my email id ? – Anjan3 Mar 8 '14 at 15:24
@AnjanDebnath, see my profile here. – lab bhattacharjee Mar 8 '14 at 16:44
@labbhattacharjee I apologize but did you provide any link in "here" ? I opened your math.stackexchange profile but could not get your contact id. – Anjan3 Mar 22 '14 at 19:08
@AnjanDebnath, – lab bhattacharjee Mar 23 '14 at 2:41
up vote 5 down vote accepted

If you need to prove that, you're in big trouble. It's Carmichael's conjecture, and it's wide open.

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Thanks for the update, actually the problem was mentioned in his "The theory of numbers" – lab bhattacharjee Aug 14 '12 at 15:58

If $n$ is odd then $\phi(n)=\phi(2n)$.

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