Determine all possible triplet(s) (E, F, G) of positive integers, with E and F being prime numbers and E ≥ F, that satisfy this equation:

E^-1 + F^-1 + (E*F) ^-1 = G^-1

(In reply to analytical solution by Daniel)

Bravo Daniel, I came to the same conclusion once my computer had found only the one triple.

My belated proof starts off like yours as far as   E + F + 1 = EF
then I rearrange to get                                   EF - E - F + 1 = 2
in order to factorise                                        (E - 1)(F - 1) = 2
Since E & F are integers with E >= F > 0,         E - 1 = 2 and F - 1 = 1, giving the triple (3, 2, 1) as the only solution.  

Posted by Harry on 2009-07-18 21:09:53