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

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SUMMARY

No Carmichael number is divisible by a perfect square greater than 1. This conclusion is based on the definition of Carmichael numbers, which are composite numbers satisfying the condition \(a^{n-1} \equiv 1 \mod n\) for all integers \(a\) coprime to \(n\). The discussion references Fermat's Little Theorem, which underpins the properties of prime numbers and extends to Carmichael numbers. The exploration of prime factors, particularly the consideration of \(p^2\) dividing a Carmichael number, is central to the proof.

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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
 
Physics news on Phys.org
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)
 

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