http://www.math.harvard.edu/~elkies/trinomial.html gives a (rational) continuum of irreducible, supposedly solvable sextics: (125 - u)*x^6 + 12*u*(u + 3)^2*x + u*(u - 5)*(u + 3)^2 but my sextic solver fails for u≠0. Can anybody propose a correction? I have no correction, but I concur with your result. For example, with u = 4 I get 121x^6 + 2352x - 196 with Galois group PSL(2,5): polgalois(121*x^6 + 2352*x - 196) %1 = [60, 1, 1, "L(6) = PSL(2,5) = A_5(6)"] which is not solvable. Charles Greathouse Analyst/Programmer Case Western Reserve University Thank you Charles! May I ask the nature of your Galwogger? I assume you observed no reducible cases for other (non0 u). If you find time, could you try one or two u and v in Matzat's allegedly solvable sextics? Not that I understand a lick of it, but this single measly bit of solvability information is the only contribution I have seen from Galois theory worth a damn to computer-algebraic denesting and solving. Has anybody yet implemented Landau's algorithm? For a ridiculous example of erudite inutility applied to solving sextics, www.sciencedirect.com/science/article/pii/S002186930098428X , which handles every possible Galois group separately, requiring 15th degree resolvents and formulæ that run on for pages on end. Excerpt: "Problem Ža. is straightforward, though with of high degree and G G48 , there are some concerns regarding memory and speed. For example, the coefficient c10 of f10Ž x. requires on the order of 9^10 arithmetic operations to be expressed. All calculations for this paper were done using Mathematica on a Sun 5.6 and this step took at most one day for any of the polynomials calculated. Solving Problem Žb. was more difficult. ..." But since the author used a computer, his worked example was printed correctly: x^6 + x^4 - x^3 - 2*x^2 + 3*x - 1 But not worked to completion, due to the godawful machinery. Anyway Ssexy got it immediately, a typical root being (1/2 (-9+Sqrt[93]))^(1/3)/(2 3^(2/3))-(1/2 (9+Sqrt[93]))^(1/3)/(2 3^(2/3))-1/(2 Sqrt[6/(-4-22 (2/(1937+201 Sqrt[93]))^(1/3)+2^(2/3) (1937+201 Sqrt[93])^(1/3))]) , an atypically simple expression. But I want simpler. The quintics that led to the denestings had as roots the sums of 5th roots of monomials. How do I find sextics and septics like that? On Mon, Aug 26, 2013 at 7:10 AM, Bill Gosper <billgosper@gmail.com> wrote: According to http://www.math.harvard.edu/~elkies/trinomial.html, modulo canonicalization, there are only five solvable, irreducible quintics with quadratic middle terms. Only one, x^5 - 100 x^2 - 1000, has roots that are the sums of fifth roots of monomials. Its real root is 2 2^(1/5) + 2 2^(2/5) - 2^(3/5) + 2^(4/5). Raising this to consecutive powers as before, In[199]:= Table[FullSimplify[Expand[%^k]], {k, 12}] Out[199]= {Root[-1000 - 100 #1^2 + #1^5 &, 1], 10 (2^(1/5) + 2^(3/5)), 10 (6 - 2 2^(1/5) + 4 2^(2/5) + 2 2^(3/5) + 2^(4/5)), 100 2^(1/5) (2 + 2^(1/5) + 2 2^(3/5)), 1000 (1 + 2^(1/5) + 2^(3/5)),[...]} The square gives this familiar-looking denesting that I can't find in my notes: In[193]:= DenestRadicals3[Sqrt[1 + 4^(1/5)]] [...] Out[193]= (-1 + 2^(1/5) + 8^(1/5) + 16^(1/5))/Sqrt[5] (Heuristic denester by Corey Ziegler Hunts.) Does anybody recognize this? I can't imagine missing it when I did sqrt(2-2^(1/7)). The fifth power gives a trinomial^(1/5) -> quadrinomial denesting In[206]:= DenestRadicals3[(1 + 2^(1/5) + 8^(1/5))^(1/5)] [...] Out[206]= (-1 + 2^(1/5) + 8^(1/5) + 16^(1/5))/125^(1/5) This I would nominate to replace the "more complicated" yet meaningless, undenestable trinomial^(1/3) inhttp://en.wikipedia.org/wiki/Nested_radical . Well, not if the article is really more about nesting than denesting. Better would be some regular polygon area formulas. But that would smack of the ever contemptible Original Research. Also missing from that article is a remark to the effect that denestings with outer radicand of the form a+b^(k/n) are quite rare when n>2. Citation needed. --rwg Actually the reason is simple: Raising a polynomial to a positive integer power hardly ever shortens it. --rwg http://www.math.harvard.edu/~elkies/trinomial.html gives a (rational) continuum of irreducible, supposedly solvable sextics: (125 - u)*x^6 + 12*u*(u + 3)^2*x + u*(u - 5)*(u + 3)^2 but my sextic solver fails for u≠0. Can anybody propose a correction? I found a djvu of B.H. Matzat: *Konstruktive Galoistheorie*, LNM 1284 (1987), where Elkies claims he got the sextic, but instead Matzat has the two families x^6 + (200000 (5 + u) (8 + u)^2 (-5 + 6 x))/((-120 + u) u^5) and x^6 - (84375 (-16 + v) v^3 (-5 + 6 x))/(4 - 14 v + v^2) Maybe one of these is Elkies's under some tricky change of variable? But woe, my sextic solver fails on these as well! Perhaps it's just broken. But it works on the cases I can find which are substantiated with actual solutions, leaving me to wonder if *both* Elkies and Matzat failed to computer-check their drafts. --rwg