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In this post, Reshetnikov gave the enormous even $80$-deg equation satisfied by,

$$x=\left|\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\color{blue}{\sqrt{\phi }}\right)\right|=1.2054797\dots\tag1$$ with golden ratio $\phi$ and absolute value $|u|$, though mentioned he was unsure if it has a solvable Galois group.

After some experimentation, it turns out an even $40$-deg equation is a satisfied by the analogous, $$y=\left|\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\,\color{blue}\phi\right)\right|=1.162132\dots\tag2$$

If we define $z=\big(\frac{4y}5\big)^2$, then we just have the $20$-deg,

$$2^{16} - 20480 z^3 - 32000 z^4 - 24576 z^5 - 25600 z^6 - 20000 z^7 - 13065 z^8 - 6000 z^9 - 309 z^{10} + 2800 z^{11} + 2500 z^{12} + 160 z^{13} + 375 z^{14} - 96 z^{15} - 100 z^{16} - 5 z^{18} + z^{20}=0$$

This is more manageable, and Magma says this has permutation group $G=2^5 \cdot3^2 \cdot 5^2 = 7200$ which is unsolvable.

Q: If $(2)$ has an unsolvable group, does that imply the same for $(1)$ as well?

Tito Piezas III
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  • Not an answer but a comment: the inner part with those fraction coefficients looks oddly like the hypergeometric function used to solve the Bring quintic form $t^5 - t - \rho = 0$. Geez, that was like 15 years ago... – The_Sympathizer Jul 30 '17 at 12:51
  • @mike4ty4: Yes, you are right. After some reflection, I actually have a good idea how to answer my own question, though I have to work out a few details. – Tito Piezas III Jul 30 '17 at 12:55
  • Then go ahead and post such an answer. If it is a good one and draws at least +3 votes we ven have a badge for it (self-learner). – Oscar Lanzi Jul 30 '17 at 13:18
  • @OscarLanzi: I've managed to reduce it from looking at the Galois group of an $80$-deg to a $24$-deg. Still high, but at least a reduction. And I'm a slow typist. :) – Tito Piezas III Jul 30 '17 at 13:44
  • @OscarLanzi: The answer is done. – Tito Piezas III Jul 30 '17 at 14:43

1 Answers1

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The scaled Bring quintic $$x^5-\frac5nx+\frac4n=0\tag1$$ is solved by, $$x =\frac45\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};n\right)$$ while the decic, $$y^4(n y^2-5) (n y^2+5)^2 + \frac{16^2}n = 16 y^3 (2 n y^2+5)\tag2$$ is solved by, $$y=\left(\frac45\right)^2\,\left|\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};n\right)\right|^2$$ with absolute value $|u|$. Eq $(2)$ then explains the $80$-deg since the argument $n=\sqrt{\phi}$ with golden ratio is a quartic root so $10\times4\times2 = 80$. Eliminating $n$ between $(1)$ and $(2)$, easily done using Mathematica, we get a solvable polynomial relationship for $x,y$, implying if one is a radical (or not), then so is the other.

However, the complete radical parameterization to $(1)$ is known and given by the Blair-Spearman quintic,

$$x^5+\frac{5u^4(4v+3)}{v^2+1}x+\frac{4u^5(2v+1)(4v+3)}{v^2+1}=0\tag3$$

Equating coefficients between $(1)$ and $(3)$, then eliminating $v$, we get the sextic in $u$,

$$n=\frac{5u^2+2u+1}{4(2-u)u^5}\tag4$$

Thus, for any radical $n$ such that $u$ is radical, then $x$ is also a radical.

Example. If $n=81$, then $u=1/3$ so we know, $$x =\frac45\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};81\right)\approx 0.42429 - 0.239932 i$$ is a radical quintic root of $x^5-\frac5{81}x+\frac4{81}=0$.

The problem then for Reshetnikov's $80$-deg is to establish if $u$ is a radical in,

$$\sqrt{\phi}=\frac{5u^2+2u+1}{4(2-u)u^5}\tag5$$

But eq $(5)$ is a $24$-deg with integer coefficients and Magma says this has permutation group $G$ of order $207360000 = 2^{12} \cdot 3^4 \cdot 5^4$. Therefore it is not solvable, hence the $80$-deg root is not a radical.

Tito Piezas III
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  • Alternatively, we can just test the Galois group of $x^5-\frac{5}{\sqrt{\phi}}x+\frac{4}{\sqrt{\phi}}=0$. This is a $20$-deg and Magma says its group $G$ has order $207360000=2^{12}\cdot3^4\cdot5^4$ and unsolvable. – Tito Piezas III Aug 01 '17 at 12:39