For what sets a,b does this theorem hold?

Quote:

Unfortunately this theorem will only hold if we have the additional constraint that divides for each prime

Just to clarify, I am not claiming that this works for *all* sets a and b. I am asking for which sets it does work. If 6k+1, 12k+1, and 18k+1 are all three primes, then their product is *always* a Carmichael. This is also true of these other two examples. I am asking for what other arithmetic progressions this happens. I doubt these are three magical ways to construct Carmichael numbers. There must be other similar arithmetic tricks. Or, perhaps these three fall into some special class of numbers that can be generalized to describe *all* Carmichaels instead of just a small subset.

There are a lot of different directions to go here...

My major short-term interest is finding a fourth or fifth construction $\displaystyle a_1k+b_1$, $\displaystyle a_2k+b_2$, $\displaystyle a_3k+b_3$ that generates Carmichaels for *all *k, with the condition that each factor is prime. Perhaps it is possible to find a finite, mutually exclusive set of generating sets for three-factor Carmichaels. If infinite, can we find a constructive way of generating all of them? The ultimate logic here is coming up with a way to *partition* the Carmichael numbers into classes based on their construction. Results about them can then be easily determined by looking at some countable list of ways they can be constructed. This is probably going to be a very difficult problem, as there are 20 million Carmichaels under $\displaystyle 10^{21}$ , ranging from 3 to 12 factors.

Let me phrase my previous question more rigorously: *Sorry, in these posts I am switching letters around.

For what sets of integers $\displaystyle a=${$\displaystyle a_1,a_2,a_3,...,a_m$} and $\displaystyle b=${$\displaystyle b_1,b_2,b_3,...,b_m$} does the following theorem hold: For all $\displaystyle k=0,1,2,... $ let $\displaystyle p_i=a_ik+b_i $ for $\displaystyle i=1,2,3,...,m$. If $\displaystyle p_i $ is prime for all $\displaystyle i$, then $\displaystyle n=p_1p_2p_3...p_m $ is a Carmichael number.

Here's a path I can see going somewhere with this problem:

1. For the three known solutions, find a common property about them, and use this property to write a single proof that these three generate Carmichaels when all factors are prime.

2. Find a few other solutions using the common properties we find from (1)

3. Try to generalize this property to describe all solutions a,b, at least for m=3, whether finite or not.

Here are some simple results, based on known properties...

Carmichaels are non-square: If $\displaystyle a_i=a_j$ , then $\displaystyle b_i \neq b_j$

$\displaystyle gcd(a_i,b_i)=1$

$\displaystyle 60k+43=6*1*(10k+7)+1$

$\displaystyle 180k+127=6*3*(10k+7)+1$

$\displaystyle 300k+211=6*5*(10k+7)+1$