# For which odd number $k$ does $\ \varphi(n) \mid n-k \$ has infinitely many solutions?

The Lehmer-totient problem: For a prime number $$\ n\$$ we have $$\ \varphi(n)=n-1\$$. In particular, we have $$\ \varphi(n) \mid n-1\$$. Is there a composite number $$\ n\$$ with $$\ \varphi(n)\mid n-1\$$ ?

It is not known if there is any such composite $$n$$ but may be Lehmer-totient problem is the tip of the iceberg. I was running computations on variation of this problem and I observed that for every small odd $$k$$ such as $$k = 3,5,7,\ldots 25$$ there were only a couple of solutions of $$\ \varphi(n) \mid n-k\$$ for $$n < 10^{10}$$. E.g. for $$k = 3$$, the only solutions so far are $$n = 9, 195$$ and $$5187$$. The data suggests that at most there are finitely many solutions for a given odd $$k$$.

Question: Is there any fixed odd $$k > 1$$ such that $$\ \varphi(n) \mid n-k\$$ has infinitely many solutions?

Conjecture: For all positive integers $$k$$ there are infinitely many $$n$$ such that $$ϕ(n)|(n − k)$$ The conjecture has been proved for $$k = 0$$, $$k = 2^a$$ for $$a> 0$$ and for $$k = 2^a\cdot 3^b$$ with $$a,b>0$$. For odd $$k\ge 3$$ the conjecture is still open.
• I don't think it's open for $k = 1$, unless the existence of infinitely many primes has come into doubt ;-) Apr 29, 2021 at 16:52