MHB Next set of prime birthdays for three brothers

[fernlund]

Hello! I have the following problem:

Three brothers are aged 6, 10 and 14 years old. Will they ever, in the future, have a prime number birthday the same year? Looking at all of the prime numbers between 1 and 100, it seems that they won't.

So I guess this is the same thing as saying: are there three consecutive prime numbers with a gap of four between the first and the second pair? Is there any way of proving this, or disproving this? Where do I start?

 

[Ackbach]

Well, to be safe, let's assume none of the brothers lives beyond 130. The primes less than 130 are
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127. Just run through them. Does the pattern you mention show up here?

 

[Sudharaka]

Hello! I have the following problem:

Three brothers are aged 6, 10 and 14 years old. Will they ever, in the future, have a prime number birthday the same year? Looking at all of the prime numbers between 1 and 100, it seems that they won't.

So I guess this is the same thing as saying: are there three consecutive prime numbers with a gap of four between the first and the second pair? Is there any way of proving this, or disproving this? Where do I start?

Hi fernlund,

This is a very interesting question. This is equivalent to asking whether in the Prime Gap sequence 4 occurs twice consecutively. I didn't find anything about the consecutive occurrences of numbers of the prime gap sequence so I suspect this might be a open problem.

 

[kaliprasad]

Forget 100 years . There is no solution
the ages are x , x+4, x+8 and working mod 3 we have 0,1,2 not necessarily in order so one of them is divisible by 3. only solution is 3,7,11 that was 3 years before and not after