Bill, Dave Dummit has a paper http://www.emba.uvm.edu/~dummit/quintics/solvable.pdf called "Solving Solvable Quintics". It appeared in Math. Comp. As far as I can see although his theorem 1 is stated for quintics with coefficients over Q, the proof works with Q replaced by any number field. Victor On Sun, Jul 24, 2011 at 2:13 AM, Bill Gosper <billgosper@gmail.com> wrote:
No replies? I expected a blizzard of unintelligible Galwology. Anyway, if the resolvent won't crack, we have the mildly surprising result that the D_5 packing of 31 disks <http://gosper.org/subopt31D5.png> is the first for which the radii are inexpressible in radicals. (Well, this may take some checking. E.g., there are D_12 and D_4 suboptimal "13"s, at least.) --rwg PS, David remarked on the absence of nondihedral rotational symmetry in his solutions. Corey remarks that he probably hasn't looked nearly far enough. For optimality in propeller type symmetry, you'd probably need >= 2 rings of disks with repeating size patterns abc and cba (or longer).
On Fri, Jul 22, 2011 at 9:06 PM, Bill Gosper <billgosper@gmail.com> wrote:
1040*(-1)^(3/5) + 912*(-1)^(7/10))*a^3 + ((-225 + 1116*I) + 2232*(-1)^(1/10) - 2964*(-1)^(3/10) + 200*(-1)^(2/5)
76142960*(-1)^(2/5) + 76142960*(-1)^(3/5) + 24743632*(-1)^(7/10))*a + ((-4765010 - 4048824*I) - 8097648*(-1)^(1/10) + 10603816*(-1)^(3/10) + 7702560*(-1)^(2/5) - 7702560*(-1)^(3/5) - 2506168*(-1)^(7/10))*a^2 + ((1630 + 336*I) + 672*(-1)^(1/10) - 1584*(-1)^(3/10) - 1040*(-1)^(2/5)
I believe the saying goes: A quintic with rational coeffs is solvable in radicals iff it factors or the Lagrange resolvent sextic (discussed here with JHC decades ago) has a rational root.
What [] if the quintic has radical coeffs? E.g., the center disk in the D_5 packing [o]f 31 disks has radius a satisfying
(43875 + 37180*I) + 74360*(-1)^(1/10) - 97460*(-1)^(3/10) - 70760*(-1)^(2/5) + 70760*(-1)^(3/5) + 23100*(-1)^(7/10) + ((47063255 + 40030096*I) + 80060192*(-1)^(1/10) - 104803824*(-1)^(3/10)
200*(-1)^(3/5) + 732*(-1)^(7/10))*a^4 - 5*a^5
Can we say that such a quintic is solvable in radicals iff the resolvent has a root expressible in radicals? If not can we at least say that no resolvent solution implies no quintic solution? -rwg I have the resolvent in Mma (LeafCount 3029) and Macsyma, if anyone wants. OLEO-GUM-RESIN NUMEROLOGIES
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