That terse solver sailed right through the examples in Quintic function - Wikipedia, the free encyclopedia<http://en.wikipedia.org/wiki/Quintic_function> until x^5 + x^4 - 4*x^3 - 3*x^2 + 3*x + 1, whose sol'ns are 2*Cos[2*k*π/11], 0<k<11. Watson-Williams's correct solution gives a quartic with four roots all of magnitude (5/Sqrt[11])^5, whereas the terse one finds four *different* roots of that magnitude, and then oviposits when it can't figure out the phase corrections. Fudged manually, the first solution gives Out[859]= Sin[7 Pi/22] == -1/10 + ( (-1)^( 2/5) (11 (-89 - 25 Sqrt[5] - 5 I Sqrt[2 (205 - 89 Sqrt[5])]))^(1/5) - (-1)^( 3/5) (11 (-89 - 25 Sqrt[5] + 5 I Sqrt[2 (205 - 89 Sqrt[5])]))^(1/5) + (-1)^(2/5) (11 (-89 + 25 Sqrt[5] - 5 I Sqrt[2 (205 + 89 Sqrt[5])]))^(1/5) - (-1)^(3/5) (11 (-89 + 25 Sqrt[5] + 5 I Sqrt[2 (205 + 89 Sqrt[5])]))^(1/5))/ (10 2^(2/5)) In[860]:= N[{%[[1]]], %[[2]], Sin[1]}, 22] Out[860]= {0.8412535328311811688618, 0.8412535328311811688618 + 0.*10^-23 I, 0.8414709848078965066525} Titus Piezas, a name I wish I'd learned much earlier, has an apparently much neater rigorous solution http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.2.3197&rep=rep1&typ... which he found by walking through a 22nd degree wall with computer algebra. It will be interesting to try on 2 cos(2k pi/11). His paper cites http://www.math.carleton.ca/~williams/papers/pdf/244.pdf but apparently slightly predates http://www.math.carleton.ca/~williams/papers/pdf/276.pdf He seems to have only one exceptional case, and not even that if you always take a limit where you might divide by 0. --rwg PS, I believe the image WFL is trying to attach is http://gosper.org/flowsnakefill_4.gif . What I'm looking for is a recursive construction of (one third of) http://gosper.org/hexflo.png (to arbitrary order) by clustering solid color tiles. PPS, there's a much higher boundary dimension variant of the France fractal, grouped three around three around one vs six around one. I'll try to unearth one.