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June 8th, 2018, 12:05 AM  #1 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Fermat Little Theorem "maximal" extension
I remember finding a paper that presented a "maximal" extension to Fermat's Little Theorem. I got as far as the case where $n$ is a product of distinct primes $p_i$ and if $l = lcm \{p_i 1 \}$ then: $$\forall x: \ x^{l+1} = x \ (mod \ n)$$ Does anyone know of an extension to this, where $n$ is any number? 
June 8th, 2018, 03:23 AM  #2 
Senior Member Joined: Oct 2009 Posts: 733 Thanks: 246  
June 8th, 2018, 04:55 AM  #3  
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Quote: It was an extension of this as well. The result I posted was for any $x$, not just $x$ coprime with $n$. It was something more general than that. It was on this site, but the older posts seem to have gone.  
June 8th, 2018, 06:41 AM  #4 
Senior Member Joined: Oct 2009 Posts: 733 Thanks: 246 
In general, you should study this through the chinese remainder theorem. This gives a ring isomorphism $\varphi$ between rings $\mathbb{Z}_n$ and $$\mathbb{Z}_{p_1^{k_1}}\times ... \mathbb{Z}_{p_l^{k_l}}$$ Thus we get $x^L = x$ if and only if $\varphi(x)^L = \varphi(x)$. And this is true if for each component in the ring $\mathbb{Z}_{p_1^{k_1}}\times ... \mathbb{Z}_{p_l^{k_l}}$ it is true. So this reduces the question to rings of the form $\mathbb{Z}_{p^k}$ which is easy. 

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extension, fermat, maximal, theorem 
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