Choose three arbitrary positive integerssuch that any two are coprime.
Define:
Lemma:See proof here: http://www.mathhelpforum.com/math-he...tml#post303617
Letand
Note that the above lemma is a necessary and sufficient condition forto exist.
Define:
Theorem: For any m such thatare all prime,
is a Carmichael Number.
Proof: Assumeare all prime. For their product to be a Carmichael Number, it is sufficient to show that
Substituting via the above construction,
Since, each of these divide the cubed term, and
divides all terms.
Thus, the sufficient condition becomes:for all
Alternatively,
Or,
Notice that by construction,and
, therefore this condition is true for all
.
Also notice that for this to hold for all,
and
. Therefore this choice for
is unique.
Q.E.D
We now have a method, given an arbitrary pairwise coprime triplet, of generating a string of Carmichael Numbers!
Results:
Forgenerates Carmichaels 1729, 294409, 56052361, 118901521 …
Forgenerates Carmichaels 29341, 1152271, 64377911, 775368901 … (For this one, k must be even, so a=30 works as a refinement)
Forgenerates Carmichaels 2465, 6189121, 19384289, 34153717249 …
Forgenerates Carmichaels 6601, 191614761361, 5124854089205401 ...


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