[Dropped eavesdroppers undropped to correctify egregium.] 4/5 3/5 2/5 1/5 1/5 2 7 - 7 - 2 7 + 6 7 + 2 sqrt(8 - 7 ) = -----------------------------------, 5 rcs>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. [...] My best guess as to what's going on is that the signs of the sqrts vary as different conjugates are selected. Bingo. Formally substituting for 7, 2 i k pi 8 i k pi 6 i k pi -------- -------- -------- 1/5 5 4/5 5 3/5 5 sqrt(8 - 7 e ) = (2 7 e - 7 e 4 i k pi 2 i k pi -------- -------- 2/5 5 1/5 5 - 2 7 e + 6 7 e + 2)/5 [k=0..4] (c42) expand(dfloat(%)) (d42) [2.55425661169849d0 = 2.55425661169849d0, 2.75838494663133 - 0.25441403376677 %i = 2.75838494663133 - 0.25441403376677 %i, 3.03551325248057 - 0.14288155378756 %i = 0.14288155378756 %i - 3.03551325248057, 0.14288155378756 %i + 3.03551325248057 = - 0.14288155378756 %i - 3.03551325248057, 0.25441403376677 %i + 2.75838494663133 = 0.25441403376677 %i + 2.75838494663133] (c43) apply("+",%) (d43) 14.1420530099223d0 = 2.0d0 Just like you said. rcs>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. No clue. In this case, the empirical truth is 4 %i %pi 2 %i %pi -------- -------- 1/5 5 1/5 5 - sqrt(8 - 7 %e ) + sqrt(8 - 7 %e ) 2 %i %pi 4 %i %pi - -------- - -------- 1/5 5 1/5 5 + sqrt(8 - 7 %e ) - sqrt(8 - 7 %e ) 1/5 + sqrt(8 - 7 ) = 2 Just like Gene said. This might be a big clue for finding denestings, if we knew how to interpret it. rcs>A likely related curiosity: 8^5 - 7 = 32761 = 181^2. YOW! And from the "+" example, (c44) sqrt(41^5*29+33^5*2^2) (d44) 59299 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., rcs>They must be huge. Huge FRAUDS! If b is the parameter, they contain both b and |b|. The good news: They remain true with |b| <- +or- b. The very bad news: When you do this, b comp[letely scales out of the expressions, leaving me with only a few dozen miscellaneous identities instead of a continuum. (But I really do have a continuum of four-termers.) --rwg PS: My pessimism about finding six-termers extends to (a^(1/n+-b^(1/n))^(1/k), k>2, not just sqrts (k=2). Of course k=-2, -1, and 2/3 constitute cheating. EAVESDROPS <-> OVERPASSED PPS: For "eavesdropper", my MW has Etymology: Middle English evesdropper, from evesdrop, n. + -er and for "eavesdrop" Etymology: probably back-formation from eavesdropper More probably than from evesdrop?