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 6k-1:

(6k-1)(2b-1)+2 = 2b(6k-1)-6k+3 
(6k-1)(2b-1)-2 = 2b(6k-1)-6k-1 
2(2b-1)+(6k-1) = 4b+6k-3 
2(2b-1)-(6k-1)  = 4b-6k-1 
2(6k-1)+(2b-1) = 2b+12k-3 
(2b-1)-2(6k-1)  = 2b-12k+1 
   
(6k+1)(2b-1)+2 = 2b(6k+1)-6k+1 
(6k+1)(2b-1)-2 = 2b(6k+1)-3(2k+1) 
2(2b-1)+(6k+1) = 4b+6k-1 
2(2b-1)-(6k+1) = 4b-6k-3 
2(6k+1)+(2b-1) = 2b+12k+1 
(2b-1)-2(6k+1) = 2b-12k-3

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 {6b-1,6b-5,4b+1,4b-5,2b+5,2b-7} all prime, and we can apply the same (6k+1), (6k-1) substitution for (2b-1) 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 2b-7=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 2012-05-20 05:15:51