The answer to the denesting of 1/5 1/5 sqrt(4 - 3 ) I mentioned last week is 3/5 4/5 3/5 2/5 2/5 4/5 1/5 1/5 - 2 3 + 3 + 2 2 3 - 2 3 + 2 ---------------------------------------------------. 5 Two days ago I found several one-parameter continua (too cumbrous to display) of solutions to sqrt(a^(1/5)+-b^(1/5)) = <five terms>, e.g., 1/5 sqrt(11 12 - 7) = sqrt(5) 3/5 4/5 1/5 3/5 4/5 2/5 2/5 1/5 - 2 3 + 2 3 - 2 3 + 4 2 3 + -------------------------------------------------, sqrt(5) 4/5 3/5 2/5 1/5 1/5 2 7 - 7 - 2 7 + 6 7 + 2 sqrt(8 - 7 ) = -----------------------------------, 5 1/5 2/5 sqrt(41 29 + 33 2 ) = 3/5 4/5 3/5 2/5 2/5 4/5 1/5 1/5 2 29 - 3 29 + 2 2 29 + 7 2 29 + 19 2 ------------------------------------------------------------, 5 4/5 1/5 sqrt(808 3 79 - 4551) = 1/5 4/5 2/5 3/5 3/5 2/5 4/5 1/5 6 3 79 - 2 3 79 + 14 3 79 - 38 3 79 - 87 ----------------------------------------------------------------. 5 However, assuming there are only finitely many rational r and d satisfying 4 3 2 2 2 0 = 2 d (2 d - 1) r + 2 d (2 d - 4 d - 1) r - (5 d - 4 d - 7) r - 2 and five close cousins, and that I have found them all, then there are no denestings of the form sqrt(a^(1/6)+-b^(1/6)) = <six real terms>, a, b positive rationals. My nearest miss: 1/6 1/3 5/6 sqrt(56 55 %i - 51) = (5 11 %i - 2 sqrt(11) %i 2/3 1/6 1/6 2/3 5/6 1/3 + 5 5 11 %i - 5 11 + 5 11 + 7 sqrt(5))/3. I'm almost ready to conjecture that there are no denestings of the form sqrt(a^(1/n)+-b^(1/n)) = <six or more real terms> for any n (with a, b positive rationals). (In spite of my sqrt(2-2^(1/7))= <five terms>.)