What is the state of Carmichael's totient function conjecture?

I have been searching for information about that conjecture and it seems for me that noone has made any significant improvement on it in the last 30 years.

Is that true? Does it remain unproven to be true? Has there been any important discovery about the problem in the last years?

Thank you.

• Are you asking about Carmichael's totient conjecture or Lehmer's? The title refers to Carmichael, but the preprint you link to is exclusively about Lehmer. These are two entirely different conjectures (Carmichael is about the multiplicity of values taken by $\phi$, and Lehmer is about the solubility of $\phi(n) \mid n-1$). – Erick Wong Oct 16 '16 at 22:17
• If you really want to know about Carmichael's conjecture, then it is obvious from the Wikipedia article that Kevin Ford has significantly advanced our understanding of it less than 20 years ago. So I repeat, do you really mean Carmichael's conjecture? – Erick Wong Feb 24 '17 at 7:16

For each $$n$$, there exists an integer $$m\neq n$$ such that $$φ(m) = φ(n) = k$$. Where $$φ$$ defined to be Euler's Totient.
The conjecture is an open problem in general, but is proven for all $$k$$ such that $$k+1$$ is prime.
Proof: Suppose $$n$$ is prime and $$n-1 = k$$. Then $$φ(n) = k$$. Now $$φ(2) = 1$$, and the totient of any integer $$t$$ is the product of totients of primes powers dividing $$t$$. Now let $$m = 2n$$. Since the only prime powers dividing $$m$$ are $$2$$ and $$n$$, $$φ(m) = (2-1)*(n-1)$$ $$=$$ $$φ(m) = k$$, therefore $$φ(m) = φ(n) = k$$.
• More generally this works for any odd $n$ - $2n$ then has the same value of the totient function. – Wojowu Feb 24 '17 at 6:34