Consider three positive integers x, y and z in geometric sequence with x < y < z < 2011 and, x+y=z+1 and determine all possible triplets (x,y,z) that satisfy the given conditions.

(In reply to Fibonacci by Math Man)

The puzzle asks for strictly increasing, so (1,1,1) is excluded.

Posted by Charlie on 2011-08-04 11:46:35