TITLE: Carmichael number square free
QUESTION [6 upvotes]: Show that if $n$ is a Carmichael number, then $n$ is a square-free.
I did this: Let $n= (p^t)(m)$  where $t >1$.
Then by modular property, 
$$b^p= b \mod n , \,\, b^m= b \mod n$$
Above two equations are also true in$\mod p$, $\mod p^t$.
But in$\mod m$, just $b^n=b \mod m$.
And I tried to use CRT
But I couldn't. 
I think that I've chosen wrong way

REPLY [4 votes]: You ask about a part of the theorem of Korselt of which I adjoint a copy. Unfortunately I don't remember the source of the publication however I give you a curious detail: Korselt has made his discovery before these numbers were called Carmichael and could not find any. It was Carmichael who discovered the first, 561,several years after this theorem.