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I know $\phi(n) = \phi(2n)$ for $n$ odd and greater than $1$. I wonder if there any value $k$ such that $\phi(n) = k$ for a unique $n$. $\phi(2) = \phi(1) = 1$ so of course $1$ cannot be that value. I know $n$ must be even.

I have looked at a few pages like this and this, but I am not sure if this question has been addressed. If there is any unique value of $\phi(n)$, then would there be infinitely many of them or only finitely many.

Josh
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    Previous discussions: https://math.stackexchange.com/questions/1971626/what-is-the-state-of-carmichaels-totient-function-conjecture also https://math.stackexchange.com/questions/181627/solutions-of-phix-n-for-a-given-n – Gerry Myerson Jun 30 '21 at 07:07

1 Answers1

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You have rediscovered Carmichael's conjecture: we believe that there is no such $k$, but it is still an open problem.

Greg Martin
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