This question is related to this but instead of $\varphi(n)$ it ask about a solution , if we replace it by the Carmichael function $\lambda(n)$
I rule out $n=1$ and $n=2$ since in this case we have $\lambda(n)=1$
Is there $n\ge 3$ besides $$230\ 608\ 999$$ such that $n^2\mid m^m-1$ holds, where $m=\lambda(n)$ and $\lambda(n)$ denotes the Carmichael function ?
As in the other problem, a solution must be odd, squarefree and cyclic. It cannot be an odd prime number. Moreover, I worked out the following :
- The above solution is the only one in the range $3\le n\le 10^9$
- Martin Hopf checked the Carmichael numbers upto $10^{18}$ (since they are cyclic). None gives a solution
- In the case of $2$ prime factors, the larger must be at least $490\ 000$.
If it is too difficult to decide whether there is a further solution, doublechecks and extensions are also welcome.