1/A = 1/B + 1/C.

One thing I noticed in your list is that A, B and C (the denominators of the original fractions) have the property that there exist positive integers a, b, and c such that A=bc; B=ac; and C=ab. Plugging these values into your first transformation of the equation we get:

BC = AC + AB => a²bc = ab²c + abc²

Dividing by abc gives us a = b + c

Some examples:



  A    B    C      a    b    c
   1    2    2      2    1    1
   2    3    6      3    2    1
   3    4   12      4    3    1
  10   14   35      7    5    2
  20   36   45      9    5    4
 330  555  814     37   22   15
 969 1330 3570     70   51   19
1189 2030 2870     70   41   29

Because of your weeding out of triples which are multiples of other triples, b and c are mutually prime in all your examples. The pattern would have been more difficult to determine and to generate without that step.

You can generate A, B, and C by

1) Set any random b' and c';
2) Set b = (b')/d; Set c = (c')/d (d = GCF(b',c')*
3) Set a = b + c
4) Set A = bcd; Set B = acd; Set C = abd

*GFC(b',c') means the Greatest Common Factor shared by b' and c'

Posted by TomM on 2003-07-20 06:58:49