Choose three arbitrary positive integers such that any two are coprime.
Lemma: See proof here: http://www.mathhelpforum.com/math-he...tml#post303617
Note that the above lemma is a necessary and sufficient condition for to exist.
Theorem: For any m such that are all prime, is a Carmichael Number.
Proof: Assume are 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
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.
We now have a method, given an arbitrary pairwise coprime triplet , of generating a string of Carmichael Numbers!
For generates Carmichaels 1729, 294409, 56052361, 118901521 …
For generates Carmichaels 29341, 1152271, 64377911, 775368901 … (For this one, k must be even, so a=30 works as a refinement)
For generates Carmichaels 2465, 6189121, 19384289, 34153717249 …
For generates Carmichaels 6601, 191614761361, 5124854089205401 ...