Is Any Carmichael Number Divisible by a Perfect Square Greater Than 1?

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Homework Statement


Prove or disprove (and salvage if possible):
No Carmichael Number is divisible by a perfect square > 1


Homework Equations


A composite number n is called a Carmichael number if and only if a^{n-1} \equiv 1 (mod \ n) for all 2\leq a \leq n-1 such that gcd(a,n) = 1

Carmichael numbers comes from fermat's little theorem that states that all prime numbers have this property. Carmichael numbers are numbers that have this property but are composite.


The Attempt at a Solution


I have been trying to figure out a way to do this problem for awhile, but no luck
 
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What if there exists a prime p such that p2 divides a?

You might want to draw some inspiration from the simplest case (what if a=p2)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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