1 Jul
2016
1 Jul
'16
9:06 a.m.
(What if we require all the numbers to be rational? I haven't figured this part out yet.)
It can be done with 4 numbers: {-2, -1/2, 1/2, 2} and cannot be done with fewer numbers. In particular, any solution with 3 rationals must necessarily correspond to a solution of the Diophantine equation: a b (a + b) = n^3 where a and b are positive integers. We can wlog assume a and b are coprime, so {a, b, a + b} are pairwise coprime. Consequently, for their product to be a cube, they must each be cubes. In particular, we obtain a positive integer solution to: x^3 + y^3 = z^3 which was proved by Gauss to have no solutions. Contradiction. Best wishes, Adam P. Goucher