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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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    @AnjanDebnath, lab.bhattacharjee@gmail.com2014-03-23

2 Answers 2

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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"2012-08-14
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If $n$ is odd then $\phi(n)=\phi(2n)$.