29 Jan
2019
29 Jan
'19
11:52 p.m.
Another tasty morsel: (1 + 3^(2/3))^2 == ((1 + 3^(1/3)) (1 + 2 3^(1/3))) (!)—rwg On Sun, Jan 27, 2019 at 9:13 AM Bill Gosper <billgosper@gmail.com> wrote:
(Empirical) claim: This identity is "primitive" or "irreducible" in the sense that it holds for no other rational exponents of the 3 and the binomials if any exponent is 0, unless they all are. —rwg
On Sat, Jan 26, 2019 at 7:54 AM Bill Gosper <billgosper@gmail.com> wrote:
3 (1 + 3 2^(1/3)) (2 - 3^(1/3)) (1 + 3^(1/3)) = = (1 + 2^(1/3))^3 (1 + 2^(2/3)) (-1 + 3^(1/3)) (-2^(2/3) + Sqrt[3]) (2^(2/3) + Sqrt[3]) (1 + 3^(2/3)) —rwg