## Posts Tagged ‘cubics’

### The equation ap^3+bq^3+cr^3+ds^3 = 0

Theorem 2:  “Likewise, given one non-trivial solution to the cubic,

$a_1y_1^3+a_2y_2^3+a_3y_3^3+\dots+a_ny_n^3 = 0$

then an infinite more can be found.”

Proof:  As before, we will start with a particular example and derive the rest by induction. The following is identically true,

$ax_1^3+bx_2^3+cx_3^3+dx_4^3 = (ay_1^3+by_2^3+cy_3^3+dy_4^3)\,(ay_1^3-by_2^3)^3$

where,

\begin{aligned} x_1 &=y_1(ay_1^3+2by_2^3)\\ x_2 &= -y_2(2ay_1^3+by_2^3)\\ x_3 &= y_3(ay_1^3-by_2^3)\\ x_4 &= y_4(ay_1^3-by_2^3) \end{aligned}

Thus, if one has initial {$y_1, y_2, y_3, y_4$} such that the RHS is zero, this leads to a second, the {$x_1, x_2, x_3, x_4$}. By iteration, these can be used to generate a third, and so on.  The reason why {$x_3, x_4$}  have a common factor will be clear in a moment.

This is by A. Desboves, but it is not hard to generalize it to n cubes. The basis is the identity,

$\text{(1)}\;\; a(ax^4+2bxy^3)^3 + b(-2ax^3y-by^4)^3 = (ax^3+by^3)(ax^3-by^3)^3$

If the first factor of the RHS, $ax^3+by^3$, can be expressed as a sum of n cubes,

$ax^3+by^3 = c_1z_1^3+c_2z_2^3+\dots +c_nz_n^3$

it is a simple matter of distributing the second factor of the RHS among the n cubes so that (1) assumes the form,

$au_1^3+bu_2^3 = c_1v_1^3+c_2v_2^3+\dots +c_nv_n^3$

which explains the common factor of {$x_3, x_4$} in the 4-cube identity.  Hence from {$x, y, z_i$}, we get new solutions {$u_i, v_i$} leading to further solutions, ad infinitum.

There doesn’t seem to be any example of a diagonal quartic surface,

$a_1y_1^4+a_2y_2^4+a_3y_3^4+\dots+a_ny_n^4 = 0$

which has been proven to have only one non-trivial and primitive integer solution.  Based upon the quadratic and cubic cases, it is tempting to speculate that if it has one, then there may be in fact an infinity.

### The equation ap^2+bq^2+cr^2+ds^2 = 0

In the previous post, it was discussed that an initial non-trivial integer solution to the diagonal quartic surface,

$ax_1^4+bx_2^4+cx_3^4+dx_4^4 = 0$

apparently does not help in determining if the equation has an infinite number of primitive integer solutions.  However, it is different for its quadratic and cubic counterparts.

Theorem 1:  “In general, given one non-trivial solution to the quadratic,

$a_1y_1^2+a_2y_2^2+\dots+a_ny_n^2 = 0$

then an infinite more can be found.”

Theorem 2:  “Likewise, given one non-trivial solution to the cubic,

$a_1y_1^3+a_2y_2^3+\dots+a_ny_n^3 = 0$

then an infinite more can be found.”

Proof of Theorem 1  (Theorem 2 will be discussed in the next post):

We’ll start with n = 3,4 and prove the rest by induction. There is the identity,

$ax_1^2+bx_2^2+cx_3^2 = (ay_1^2+by_2^2+cy_3^2)\,(az_1^2+bz_2^2+cz_3^2)^2$

where,

$(x_1, x_2, x_3) = (uy_1-vz_1, uy_2-vz_2, uy_3-vz_3)$

and,

$(u,v) = (az_1^2+bz_2^2+cz_3^2,\, 2(ay_1z_1+by_2z_2+cy_3z_3)\,)$

Thus, if one has an initial solution $(y_1, y_2, y_3)$, then the RHS of the identity becomes zero, and one gets a parametrization in the $(x_1, x_2, x_3)$ for three free variables $(z_1, z_2, z_3)$. An example is given in this page, form 2b.  For n = 4, it is just a generalization,

$ax_1^2+bx_2^2+cx_3^2+dx_4^2 = (ay_1^2+by_2^2+cy_3^2+dy_4^2)(az_1^2+bz_2^2+cz_3^2+dz_4^2)^2$

where,

$(x_1, x_2, x_3, x_4) = (uy_1-vz_1, uy_2-vz_2, uy_3-vz_3, uy_4-vz_4)$

and,

$(u,v) = (az_1^2+bz_2^2+cz_3^2+dz_4^2,\, 2(ay_1z_1+by_2z_2+cy_3z_3+dy_4z_4)\,)$

The pattern is easily seen for n = 5,6, ad infinitum.  If only it was that easy for the quartic case.

### 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.

### Cubic equations

Recall that solving the quadratic equation,

$x^2+ax+b = 0$

gives,

$x = \frac{1}{2}\,(-a \pm \sqrt{y}\,)$

where y is the linear,

$y = a^2-4b$

The solution to the cubic equation is simply a generalization. Given,

$x^3+ax^2+bx+c=0$

then,

$x =\frac{1}{3}\,(-a+y_1^{1/3}+y_2^{1/3})$

where the $y_i$ are now the two roots of the quadratic,

$y^2+(2a^3-9ab+27c)y+(a^2-3b)^3 = 0$

Notice that the constant term of its resolvent is a perfect cube.  For prime degrees, if the irreducible equation is solvable, then it can be solved analogously. For depressed quintics (an easy linear transformation to make a = 0), then,

$x = \frac{1}{5}\,(y_1^{1/5}+y_2^{1/5}+y_3^{1/5}+y_4^{1/5})$

where the $y_i$ are now the four roots of its resolvent quartic. For non-binomial quintics, this quartic will have a constant term that is a perfect fifth power. And so on for other prime degrees.

### Solvable sextics

In a previous post, we discussed the Bring-Jerrard quintic,

$x^5+ax+b=0$

and rational parametrizations for {a, b}.  It turns out this has a sextic version. To illustrate, these irreducible equations are solvable,

$\text{(1)}\;\; x^6+3x+3 = 0$

$\text{(2)}\;\; x^6-7^2x-7^2 = 0$

However, each needs a different approach. The first factors into two cubics over $\sqrt{-3}$, while the second factors into three quadratics whose coefficients are determined by a cubic.  These two examples belong to two infinite (but not complete) families.

(1)  1st family:

$x^6+4mn(m^2-3n)x+n(m^4+18m^2n+n^2) = 0$

This factors over $\sqrt{-n}$.  Let {m, n} = {-1/2, 3/4}, and it will yield (1).

(2)  2nd family:

The second is trickier.  Given,

$x^2+vx+(pv^2-qv-nr) = 0$

$v^3-rv-r = 0$

Eliminating v between them using resultants (easily done in Mathematica or Maple), results in a sextic in the variable x of form,

$x^6+ax+b = 0$

if,

\begin{aligned}p &=\frac{3n+1}{2}\\q &=\frac{9n+1}{4}\\r &= \frac{(9n+1)(9n-7)}{4(3n+1)(n-1)} \end{aligned}

Let n = 3, we get {p, q, r} = {5, 7, 7}, and it yields (2).  Thus, to solve,

$x^6-7^2x-7^2 = 0$

entails solving the cubic in v,

$v^3-7v-7 = 0$

though other n do not show this nice symmetry.  Another good value is at n = 1/3 which gives,

$x^6+3x+5 = 0$