rwg writes ... 4/5 3/5 2/5 1/5 1/5 2 7 - 7 - 2 7 + 6 7 + 2 sqrt(8 - 7 ) = -----------------------------------, 5 This set off a "can't be right" buzzer for me. Since it's an algebraic identity, it should also be valid for the complex 5th roots of 7. Suppose we add up all five conjugates. As we run through the 5 fifth roots of 7, the LHS radicand will have real parts between 6 and 10, and we can expect the sum of all five sqrts to be roughly 5*sqrt8, about 14. But in summing the RHS, all the 7^(k/5) terms will cancel, since the sum of the five fifth roots of 7 is 0. This leaves only the 2/5 term, times 5, = 2. Thus, an apparent contradiction, 14 = 2. So I checked the equation numerically, and it's correct in double precision, about 16 decimal places. PRIME>(sqrt (- 8 (expt 7 1/5))) 2.5542566116984893 PRIME>(/ (+ (* 2 (expt 7 4/5)) (* -1 (expt 7 3/5)) (* -2 (expt 7 2/5)) (* 6 (expt 7 1/5)) 2) 5) 2.5542566116984893 My best guess as to what's going on is that the signs of the sqrts vary as different conjugates are selected. But this creates two new puzzles: (a) There are only a few combinations of sign assignments that will still cancel the imaginary parts when summing the LHS, and it's mysterious that the real parts mostly cancel too. (b) And, what's the rule or rationale for selecting the LHS sqrt signs? Normally one expects equations with different (relatively prime) degree roots to permit separate choices of the roots. A likely related curiosity: 8^5 - 7 = 32761 = 181^2. rwg> 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., They must be huge. Rich rcs@cs.arizona.edu