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.


1 Answer 1


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

  • $\begingroup$ Carmichael proposed this as an exercise in his 1914 intro Number Theory textbook. $\endgroup$ Jun 30, 2021 at 7:04
  • $\begingroup$ @GerryMyerson Does that mean he knew the proof? $\endgroup$
    – Asher2211
    Jun 30, 2021 at 7:13
  • 2
    $\begingroup$ @Asher, it means he thought he knew a proof. He was wrong. See also the link Greg provided. $\endgroup$ Jun 30, 2021 at 7:15
  • 3
    $\begingroup$ The article mentions the incredible lower bound for a possible solution : It is $10^{10^{10}}$ (!!) $\endgroup$
    – Peter
    Jun 30, 2021 at 9:03
  • $\begingroup$ @Peter: Was that lower bound obtained via analytical methods? $\endgroup$ Jul 6, 2021 at 3:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .