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,


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 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


\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


\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.