Find the number of triplets of prime numbers in ascending order such that the difference between each pair of successive numbers is 2. (a.2 b.3 c.1 d.4)?

Question

Find the number of triplets of prime numbers in ascending order such that the difference between each pair of successive numbers is 2. (a.2  b.3  c.1  d.4)?

kainui · Answer

Well if we were to construct them, we'd have this as the general form of our triplets:

$$(p, p+2, p+4)$$

It doesn't take too long before primes are only of the form of $6n-1$ and $6n+1$ which leaves out $6n-3$ and $6n+3$ since these are divisible by 3, so they can't be prime at all.

If you're unfamiliar with what I mean, 6n, 6n+1, 6n+2, 6n+3, 6n+4, and 6n+5 can represent every number much like 2n and 2n+1 can represent every number by splitting them up into the even and odd numbers.

debpriya · Answer

so basically when we try this we get 6n+3 in every triplet ....therefore there is no such triplet ? what is the exact answer?

kainui · Answer

Almost, however this divisibility thing doesn't work for when n=0.