Yeah, I've already found it. But I'm rather asking "what is equivariant homology ?".

Huuu... This is not encouraging, especially since I don't know what a Mackey functor and most of those concepts are...

So you add a constraint ($a \equiv 1 (n/p^k)$) to find an element. Quite clever, I didn't see that one. Thank you very much !

Thank you very, I begin to see the pattern of the proof. Though I do not need Korselt Criterion, I shall prove it on the run, so it's done. But one big problem remains for me : why do $(a,n)=1$ ? for instance if $n$ would be $p^k(1+p^{k-1})$, the statement would be wrong since $a|n$. Of course such a $n$ is not a caermichael number for $k \ge 2$ since it is not square-free, but that's what I want to prove ! Finding coprimes is a recurring issue I cannot get rid of. PS: Sorry for the answer, I use another account to answer. I didn't understand you can recover an account.