Very very nice! Jim Propp On Friday, July 1, 2016, Adam P. Goucher <apgoucher@gmx.com> wrote:
(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
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