On Mon, Sep 2, 2013 at 2:51 PM, Noam Elkies <elkies@math.harvard.edu> wrote:
Bill Gosper <billgosper@gmail.com> writes:
Our sextic solver fervently assumes (I think with Piezas's blessing) that the solvable ones all factor into either cubics with quadratic surd coefficients, or vice versa.
That's right, at least if "cubic surd" means the solution of an arbitrary cubic, not just expressions rational in n^(1/3).
NDE>Are you suggesting there's some significance to the large convergents?
Well, it can't hold a candle to Brillhart's In[228]:= ContinuedFraction[(15 Sqrt[3]-Sqrt[163])^(1/3)/Sqrt[3]+(15 Sqrt[3]+Sqrt[163])^(1/3)/Sqrt[3],239] [...] Hmm, \sqrt{163}.)
I don't know whether the article is online, but I remember the result, and yes, it certainly has to do with sqrt(163) yoga. Even if it "doesn't hold a candle" to Brillhart's cubic, it might point the way to other sextics whose roots have more remarkable rational approximations.
Oops, Piezas (https://sites.google.com/site/tpiezas/001) gives the sextic 1 - 10 x^3 - 640320 x^5 + 5 x^6 with a root (also of the quadratic 264840 + 87 10^( 2/3) ((282094543 - 21 Sqrt[489])^( 1/3) + (282094543 + 21 Sqrt[489])^(1/3)) - 15 (256128 + 2^(2/3) Sqrt[ 3] (5 (93346861716363 Sqrt[3] - 54578363 Sqrt[163]))^(1/3) + 2^(2/3) Sqrt[ 3] (5 (93346861716363 Sqrt[3] + 54578363 Sqrt[163]))^(1/3)) x + 90 x^2) with CF = {128064, 8200194048, 62470, 4, 10, 6513259, 1, 2, 28, 1, 1, 2, 2, 4, 2, 1, 3, 9, 1, 329, 2, 2, 3, 1, 1, 1, 5, 2, 2, 2,...} (489=3*163) Are there more spectacular e^(π algebraic)?
NDE> No septics here. This family of octics factors over quadratic NDE> extensions, so it's "just" solving a quartic over Q(sqrt(d)).
Yow, it Gaussian-factors!
Right, in this case d=-1. But you'd still have to go through the solution of a quartic by radicals to fully write out a root this way. Likewise for the mutant trinomial x^8 + 4*x^5 + 8 -- and indeed this one also "Gaussian-factors" (I exhibited it as A^2+B^2), though it's unique up to scaling even without specifying the quadratic extension over which it factors into two quartics.
The curves parametrizing octic trinomials with Galois group contained in (2^3).7 (the ax+b group over the 8-element field), or even ((2^3).7).3 (extending by the field automorphism), are much more complicated, and I wouldn't be surprised if there are no non-degenerate rational points.
NDE
P.S. Are all of those e-mails being seen on math-fun too? I'm not a subcriber and keep getting auto-replies "Your mail to 'math-fun' with the subject <...> Is being held until the list moderator can review it for approval. The reason it is being held: Post by non-member to a members-only list".
They eventually came out in a bunch. People will think you're manic. --rwg