Unsolved Problem Q103001



Carmichael's totient function conjecture


Statement:

For every positive integer \(n\) there exists an \(m \neq n\) such that \( \phi(m)=\phi(n) \)

Author  Date Published 
     
Robert Carmichael  1907 


Partial Results:

Positive integers \(n\) such that \(\phi(n)=k\)  OEIS: A032447

Number of positive integers \(n\) such that \(\phi(n)=k\)  OEIS: A014197

A lower bound was given by Kevin Ford in 1998 [1] that states if a counter example exists it must be larger than \(10^{10^{10}}\)

Pomerance in 1974 [2] gave a highly improbable sufficient condition: \(n\) is a counterexample if for every prime \(p\) such that \(p-1\) divides \(\phi(n) \) then \(p^2\) divides \(n\).



Related Problems:


References:

[1] Ford, K. (1999), "The number of solutions of φ(x) = m", Annals of Mathematics, 150 (1): 283–311, doi:10.2307/121103, JSTOR 121103, MR 1715326, Zbl 0978.11053.


[2] Pomerance, Carl (1974), "On Carmichael's conjecture" (PDF), Proceedings of the American Mathematical Society, 43 (2): 297–298, doi:10.2307/2038881, JSTOR 2038881, Zbl 0254.10009.



Keywords:

Analytic Number Theory; Carmichael's totient function conjecture