## Posts Tagged ‘octics’

### Quartic equations

Composite degrees n are better solved by polynomial decomposition.  However, for $n = 2^m$, there is an alternative method.  Thus, we’ll give two methods to solve the quartic.  Given the depressed quartic,

$x^4+px^2+qx+r = 0$

Its resolvent cubic is simply,

$y^3+2py^2+(p^2-4r)y-q^2=0$

Method 1:

Given any root y of the cubic, then,

\begin{aligned} x_{1,2} &=\tfrac{1}{2} \left(\sqrt{y} \pm \sqrt{-(2p+y)-\tfrac{2q}{\sqrt{y}}}\,\right)\\ x_{3,4} &= \tfrac{1}{2} \left(-\sqrt{y} \pm \sqrt{-(2p+y)+\tfrac{2q}{\sqrt{y}}}\, \right)\end{aligned}

For convenience, one can choose the real root y.  Notice that {$x_3, x_4$} are the negative case of $\pm \sqrt{y}$.  The formula can easily be proven by squaring the variable y, forming the quadratic in {$x_1,\,x_2$}, and eliminating y between them using resultants,

$\text{Factor}[\text{Resultant}[y^6+2py^4+(p^2-4r)y^2-q^2,\,x^2+xy+\frac{y^3+py-q}{2y},\,y]]\\ = (x^4+px^2+qx+r)^3$

proving any root of the sextic (a cubic in $y^2$) will suffice.

Method 2:

Given the three roots {$y_1, y_2, y_3$} of the resolvent cubic, then,

\begin{aligned} x_{1,2} &= \frac{\sqrt{y_1} \pm (\sqrt{y_2}+\sqrt{y_3}\,)}{2}\\ x_{3,4} &= \frac{-\sqrt{y_1} \pm (\sqrt{y_2}-\sqrt{y_3}\,)}{2}\end{aligned}

where signs are chosen such that,

[1]  $\sqrt{y_1}\sqrt{y_2}\sqrt{y_3} = -q$

Proof:  Expanding,

$(x-x_1)(x-x_2)(x-x_3)(x-x_4) = 0$

using the formulas, we get,

$x^4-\frac{1}{2}(y_1+y_2+y_3)x^2-\sqrt{y_1}\sqrt{y_2}\sqrt{y_3}\,x\\+\frac{1}{16}(y_1^2+y_2^2+y_3^2-2(y_1y_2+y_1y_3+y_2y_3)) = 0$

From [1] and the properties of elementary symmetric polynomials, we find that it is in fact equal to,

$x^4+px^2+qx+r = 0$

A similar method was used to solve the octic $x^8-x^7+29x^2+29 = 0$ in this post, though now one had to take square roots of its resolvent septic.

### 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.