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

(1/e)+(1/f)+(1/ef)=1/g
(e+f+1)/(ef)=1/g
g=ef/(e+f+1)
only way g can be an integer is if e+f+1 is either e,f, or ef because e,f are prime.

Only one that is possible is e+f+1=ef thus
ef-e=f+1
e(f-1)=f+1
e=(f+1)/(f-1)
since e>=f we have
(f+1)/(f-1)>=f
f+1>=f(f-1)
f+1>=f^2-f
f^2-2f-1<=0
thus
1-sqrt(2)<=f<=1+sqrt(2)
-0.4<=f<=2.4
only prime that f can be then is 2
thus f=2
e=(2+1)/(2-1)=3/1=3
and
g=(2*3)/(2+3+1)=6/6=1

thus the only possible triplet is (3,2,1)

Posted by Daniel on 2009-07-18 13:20:06