With reference to the Euler's totient function $\phi(\cdot)$, given any $n \in \mathbb{Z}^+$, it's quite straightforward to find $\phi(n)$.

In contrast, given $n \in \mathbb{Z}^+$, even though there are to find the $k \in \mathbb{Z}^+$ such that $\phi(k) = n$, I'm not aware of any method to determine the number of such $k$ values beforehand. What are the progress that has been made in this regard ?

  • $\begingroup$ Number of numbers $m$ with Euler $\phi(m) = n$ is tabulated at oeis.org/A014197 $\endgroup$ Apr 18 at 0:05
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    $\begingroup$ Cf. this question $\endgroup$ Apr 18 at 0:06
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    $\begingroup$ Actually, your first sentence is not really true if by "straightforward" you mean "reasonably easy to compute." If, for example, $p$ and $q$ are thousand digit primes and $n=pq$, then it is currently completely infeasible to compute $\phi(n)$ (unless one has a quantum computer in one's basement!). The point is that computing $\phi(n)$ is equivalent to finding $p$ and $q$. RSA, and many other cryptographic algorithms, depend on the difficulty of this problem. However, I realize this isn't the gist of your question, which is interesting. $\endgroup$ Apr 19 at 22:26
  • $\begingroup$ @JoeSilverman, thank you for the comment, Professor ! You're one of my mathematical heroes. $\endgroup$
    – Eureka
    Apr 20 at 5:59

2 Answers 2


Ford's theorem ($1999$) states that for any $ m\ge2,$ there exists a totient number $n$ with multiplicity $m$ (that's for which there are $m$ solutions to $\varphi (x)=n.$)* This had been conjectured by Sierpiński.

Also, each multiplicity occurs infinitely often.

Furthermore, no number is known with multiplicity $1.$ Carmichael conjectured that there's no such number.

*See The Number of Solutions of $\varphi (x)=m,$ Annals of Mathematics, Second Series, Vol. 150, No. 1, Jul 1999, pp. 283-311 (29 pages)


You may like to check my paper "Computing the Inverses, their Power Sums, and Extrema for Euler's Totient and Other Multiplicative Functions".

There is also a PARI/GP implementation of the proposed algorithm.

  • $\begingroup$ Thank you ! The content of the paper is too abstract to grasp. I'll check it anyways. $\endgroup$
    – Eureka
    Apr 18 at 2:48

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