I was reading up on Carmichael numbers and I came across a brief proof that numbers of the form,(adsbygoogle = window.adsbygoogle || []).push({});

[tex]n = (6m+1)(12m+1)(18m+1)[/tex]

where [itex]m \in \mathbb{Z}[/itex] for which [itex]6m+1[/itex], [itex]12m+1[/itex], and [itex]18m+1[/itex] are all prime, are all Carmichael numbers.

A quick search on this proof directed me to many of the various math reference sites and showed a number of properties which leads to the conclusion that n is a Carmichael number. These include

[1] [itex]36m[/itex] is a multiple of [itex]n-1[/itex] - which I proved easily via long division.

[2] The lowest common multiple of [itex]6m[/itex], [itex]12m[/itex], [itex]18m[/itex] is [itex]36m[/itex] - which surprisingly corresponds to [1].

[3] Therefore since [itex]36m[/itex] divides [itex]n[/itex] and is the least common multiple of 6m, 12m, and 18m then for any integer a coprime to n,

[tex]a^{n-1} \equiv 1(\mod 6m+1)[/tex]

which is just a simple reverse application of Fermat's Little Theorem. The same can be said of 12m and 18m:

[tex]a^{n-1} \equiv 1(\mod 12m+1)[/tex]

[tex]a^{n-1} \equiv 1(\mod 18m+1)[/tex]

[4] Therefore by a Corollary of F.L.T. we have

[tex]a^{n-1} \equiv 1(\mod n)[/tex]

So for any integer a coprime n. Which is exactly the definition of a Carmichael number.

Have I got all 4 points correct?

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# Homework Help: I do not understand this proof

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