The prime triplet (2, 3, 13) is interesting in that it can produce six primes using a product with a sum or difference:
2 + 3*13 = 41
3 + 2*13 = 29
13 + 2*3 = 19
3*13  2 = 37
2*13  3 = 23
13  2*3 = 7
Prove this is the only such prime triplet.
We can see that 2 must be one of the 3 primes, and that there can't be two 2's.
The key to the problem is that 3 must be another: assume by way of contradiction that the smaller of the two remaining primes is greater than 3. Then, using the result that all primes greater than 3 are of the form 6k+1 or 6k1:
(6k1)(2b1)+2 = 2b(6k1)6k+3
(6k1)(2b1)2 = 2b(6k1)6k1
2(2b1)+(6k1) = 4b+6k3
2(2b1)(6k1) = 4b6k1
2(6k1)+(2b1) = 2b+12k3
(2b1)2(6k1) = 2b12k+1
(6k+1)(2b1)+2 = 2b(6k+1)6k+1
(6k+1)(2b1)2 = 2b(6k+1)3(2k+1)
2(2b1)+(6k+1) = 4b+6k1
2(2b1)(6k+1) = 4b6k3
2(6k+1)+(2b1) = 2b+12k+1
(2b1)2(6k+1) = 2b12k3
and by inspection, not all of those in either group can be prime at the same time.
So now we have 2 and 3 as two of the primes (and the third cannot be 3), with {6b1,6b5,4b+1,4b5,2b+5,2b7} all prime, and we can apply the same (6k+1), (6k1) substitution for (2b1) also. Although I haven't worked this through fully since it seems quite a dry exercise, I expect the only possible compliant result is that 2b7=b, so that b=7 and the third member of the triplet is accordingly 13.
Note: Having read Mathman's solution after doing some of the workings myself, I believe that we are thinking along essentially the same lines.
Edited on May 20, 2012, 5:37 am

Posted by broll
on 20120520 05:15:51 