## Posts Tagged ‘septics’

### A Tale of Three Solvable Octics

The following three cute irreducible octics are solvable,

$\text{(1)}\;\; x^8-5x-5 = 0$

$\text{(2)}\;\; x^8-44x-33 = 0$

$\text{(3)}\;\; x^8-x^7+29x^2+29 = 0$

However, each needs to be solved in a different way: they need a quadratic, quartic, and septic subfield, respectively.

The first is the easiest, it factors over $\sqrt{5}$ into two quartics.  The second does not factor over a square root extension, but factors into four quadratics,

$x^2+vx -(2v^3-7v^2+5v+33)/13 = 0$

where the coefficients are determined by the quartic,

$v^4+22v+22 = 0$

The command,

Resultant[x^2+vx -(2v^3-7v^2+5v+33)/13 , v^4+22v+22 ,v]

done in Mathematica or in www.wolframalpha.com will eliminate the variable v and recover [2].  The first two octics are by this author.  (Any other example for the second kind with small coefficients?)

The third (by Igor Schein) is the hardest, as it needs a septic subfield. Interestingly though, the solution involves the 29th root of unity. Given,

$x^8-x^7+29x^2+29 = 0$

Then,

\begin{aligned} x_1 &= (1+(a-b-c-d+e-f-g))/8\\ x_2 &= (1-(a-b-c-d-e+f+g))/8\\ x_3&= (1-(a+b-c+d+e-f+g))/8\\ x_4&= (1+(a+b-c+d-e+f-g))/8\\ x_5&= (1-(a+b+c-d+e+f-g))/8\\ x_6&= (1+(a+b+c-d-e-f+g))/8\\ x_7&= (1-(a-b+c+d-e-f-g))/8\\ x_8&= (1+(a-b+c+d+e+f+g))/8 \end{aligned}

where the 7 constants {a, b, c, d, e, f, g} is the square root of the appropriate root $z_i$ of the solvable septic,

$z^7-7z^6-2763z^5-19523z^4+1946979z^3+34928043z^2+\\119557031z-3247^2 = 0$

namely,

\begin{aligned} a &\approx \sqrt{ -26.98}\\ b &\approx \sqrt{ -26.95}\\ c &\approx \sqrt{ -19.71}\\ d &\approx \sqrt{ -4.78}\\ e &\approx \sqrt{ 0.08}\\ f &\approx \sqrt{ 36.91}\\ g &\approx \sqrt{ 48.43}\\ \end{aligned}

Note that,

$(8x_3-1)+(8x_4-1)+(8x_5-1)+(8x_6-1) = -4e$

hence one can use the roots of this octic to express the roots of its resolvent septic, and vice versa.  In terms of radicals, Peter Montgomery expressed the septic roots $z_i$ as,

\begin{aligned}\tfrac{1}{4}(z-1) &= 2(w^{11}+w^{13}+w^{16}+w^{18})-2(w+w^{12}+w^{17}+w^{28})\\ &+(w^3+w^7+w^{22}+w^{26})-(w^2+w^5+w^{24}+w^{27})\\ &+(w^4+w^{10}+w^{19}+w^{25})-(w^8+w^9+w^{20}+w^{21}) \end{aligned}

where w is any complex root of unity (excluding w = 1) such that $w^{29}=1$. For example, $w = \exp(2\pi i\cdot4/29)$ will yield the value for $z_1 \approx -26.98$,  and so on.

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### A Family of Solvable Quintics and Septics

Define,

$x = \frac{-\sqrt{2}\,\eta(2\tau)}{\zeta_{48}\,\eta(\tau)}$

where $\eta$ is the Dedekind eta function, and $\zeta_{48}$ is the 48th root of unity.  Then for $\tau = \frac{1+\sqrt{-d}}{2}$ for d = {47, 103}, x is a root of the quintics,

$x^5-2x^4+2x^3-x^2+1 = 0$

$x^5-2x^4+3x^3-3x^2+x+1 = 0$

respectively. Note that the class number h(d) of both is 5.  It turns out these belong to a family of solvable quintics found by Kondo and Brumer,

$x^5-2x^4+2x^3-x^2+1 = nx(x-1)^2$

for any n, and where the two examples are n = {0, -1}.  A similar one for septics can be deduced from the examples in Kluner’s A Database For Number Fields as,

$x^7-2x^6+x^5-x^4-5x^2-6x-4 = n(x-1)x^2(x+1)^2$

with discriminant,

$d = 4^4(4n^3+99n^2+34n+467)^3$ .

The case n = 0 implies d = 467 and, perhaps not surprisingly, the class number of h(-467) = 7. However, since 467 does not have form 8m+7, then the eta quotient will be not be an algebraic number of degree h(-d).

To find a solvable family, it’s almost as if all you need is to find one right solvable equation, affix the right n-multiple of a polynomial on the RHS, and the whole family will remain solvable.

### Solvable quintics

Here is a nifty sufficient but not necessary condition on whether a quintic is solvable in radicals or not.  Given,

[1]  $x^5+10cx^3+10dx^2+5ex+f = 0$

If there is an ordering of its roots such that,

[2]  $x_1 x_2 + x_2 x_3 + x_3 x_4 + x_4 x_5 + x_5 x_1 - \\(x_1 x_3 + x_3 x_5 + x_5 x_2 + x_2 x_4 + x_4 x_1) = 0$

or alternatively, its coefficients are related by,

[3]  $-25c^6-40c^3d^2-16d^4+35c^4e+\\28cd^2 e -11c^2e^2+e^3-2c^2df-2def+cf^2 = 0$

then [1] is solvable as,

$x = z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}$

where the $z_i$ are the roots of the simple quartic,

$z^4+fz^3+(2c^5-5c^3e-4d^2e+ce^2+2cdf)z^2-c^5fz+c^{10} = 0$

Note that [3] in fact is the constant term of Cayley’s resolvent sextic and is only quadratic in f.  Using another relation among the $x_i$, Dummit’s resolvent has a constant term that is already a quartic in f, hence the choice of relation matters.

Example 1: This family of quintics by this author satisfies [3],

$x^5+10x^3+5(n^2+3n+18)x^2-5(n^3+n^2+15n-14)x+\\(n^4-n^3+37n^2+441) = 0$

Let n = 1, and we have,

$x^5+10x^3+110x^2-15x+478 = 0$

Quartic is,

$z^4+478z^3+11994z^2-478z+1 = 0$

such that,

$x = z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5} = -4.50991\dots$

Example 2: Another good example of [3] is Emma Lehmer’s quintic,

$y^5 +n^2y^4-(2n^3+6n^2+10n+10)y^3+\\(n^4+5n^3+11n^2+15n+5)y^2+(n^3+4n^2+10n+10)y+1$

The linear transformation,

$y = x-n^2/5$

will reduce it into the form of [1], and it will then be seen its coefficients obey [3]. As a particular example, let n = 5 and we have the reduced form,

$x^5-710x^3+11005x^2-59640x+108701 = 0$

Let its roots be,

$(x_1, x_2, x_3, x_4, x_5) \approx (-33.15,\; 4.83,\; 12.16,\; 4.99,\; 11.16)$

and we find that indeed it obeys [2].

Unfortunately, no similar simple relation between the coefficients of a solvable septic, or 7th degree equation, is yet known.