That is just for primes up to 71. Looks like there is an infinite number. I can see no reason why it should stop.

phanthanhtom

2013-06-16 14:11:10

One more problem: I have checked the prime numbers table <6000. The number of prime quadruplets are 9 (excluding 5-7-11-13), and the number of near-quadruplets (i.e. exactly one number missing, if this number is added it makes a 10k+1, 10k+3, 10k+7 and 10k+9 quadruplet) is 37. 37/9=4.25 rounds to 4 = 4C3 (combination without repetition). Interesting!

phanthanhtom

2013-06-15 21:23:22

There might be an infinite number of solutions. I think so. And there might be two pairs of the same numbers. Or a = b and c = d. As long as not all of them are the same. (In fact I acknowledge that that sentence is redundant)

bobbym

2013-06-15 19:55:30

Hi phanthanhtom;

First thing you have to do is change this:

because of this

Find all sets of 4 different prime numbers (a, b, c, d)

And I am finding many solutions. Perhaps there are an infinite number. Or perhaps something is wrong in my method.

phanthanhtom

2013-06-15 18:38:51

Find all sets of 4 different prime numbers (a, b, c, d) with

such that the sum of any three out of these four primes are also prime. In other words, find positive integers a, b, c, d such that a, b, c, d, a+b+c, a+b+d, a+c+d and b+c+d are eight different primes.