My Math Forum Fermat Little Theorem "maximal" extension

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 June 7th, 2018, 11:05 PM #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, 02:23 AM #2 Senior Member   Joined: Oct 2009 Posts: 544 Thanks: 174
June 8th, 2018, 03:55 AM   #3
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Quote:
 Originally Posted by Micrm@ss https://en.wikipedia.org/wiki/Carmichael_function
Thanks.

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, 05:41 AM #4 Senior Member   Joined: Oct 2009 Posts: 544 Thanks: 174 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. Thanks from Pero

 Tags extension, fermat, maximal, theorem

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