Carmichael number square free 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
 A: We start as in your post, letting $n=p^tm$ where $t\ge 2$ and $m$ is not divisible by $p$. 
By the Chinese Remainder Theorem, the system of congruences $x\equiv 1+p\pmod{p^t}$, $x\equiv 1\pmod{m}$ has a solution $a$. Note that $\gcd(a,n)=1$.
Since $n$ is Carmichael, we have $a^{n-1}\equiv 1\pmod{n}$. In particular, $a^{n-1}\equiv 1\pmod{p^t}$, and therefore $a^{n}\equiv a\pmod{p^2}$.
So $(1+p)^{n}\equiv 1+p\pmod{p^2}$. Expand $(1+p)^{n}$ modulo $p^2$ using the binomial theorem. We get that $(1+p)^{n}\equiv 1\pmod{p^2}$, since the first two terms of the expansion are $1$ and $np$, and the rest of the terms are divisible by $p^2$.
Thus $1\equiv 1+p\pmod{p^2}$. This is impossible.
A: 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.

