Are there four numbers in AP such that their prime factors are also in AP? The prime factors of $40807$ are $(13, 43, 73)$ are in arithmetic progression.
The prime factors of $55125 $ are $(3, 5, 7)$ are also in AP.
The prime factors of $69443$ are $(11, 59, 107)$ are also in AP.
Moreover, the three numbers $(40807, 55125, 69443)$ are also in AP. It is the smallest AP of three positive composite integers $(a,b,c)$ such that $\gcd(a,b,c) = 1$ and the prime factors of each of these numbers are also in some arithmetic progression. There are several other such triplets. But when it comes to four such numbers, there seems to be none. Is there any reason why a such a AP of four numbers cannot exist?
Question 1: Is there an AP of four composite coprime positive integers such that the prime factors of each of these four numbers are in some AP?
Update: No solutions below $2.5 \times 10^9$.
 A: First find three or more primes in arithmetic progression. You can easily show that the first prime is not 2; if the first prime is 3 then there are no more than three numbers in the progression because the fourth number is divisible by 3; if the first prime is >= 5 then the difference between the primes is a multiple of 6, otherwise one of the next two numbers is divisible by 3.
Then count the number of integers $2^{k-1} <= N < 2^k$ where N has three or more distinct prime factors in arithmetic progression, say for k <= 60. You do this by creating a table of primes, finding all primes p, q, r in arithmetic progression with pqr < 2^k from that table (that’s fast, because r = 2q - p, and pqr < 2^k limits the possibilities), then adding pqr to the table, possibly multiplied by p, q, r. And of course the same for more primes in arithmetic progression. But there will be few since their product is limited.
And then you look at your numbers and estimate the chances to find four numbers in arithmetic progression in the table. That chance will be quite small for k = 60. Then you see how these chances change with k, and if the shrink quickly with k, then finding four numbers will be unlikely.
That doesn’t mean it’s impossible, but it might be.
PS. The table of primes you need is not large, the largest r would happen if 3p(2p-3) < 2^k, so the largest r is less than sqrt(2^k * 2/3).
