In this section we describe some "unexpected" symmetry in algebraic identities derived from cubic and quartic curves. While these identities can be proved by factorising each side into products of Weierstrass $\sigma$-functions using elliptic function theory, they can also be obtained from simple geometric constructions and then proved using direct algebraic computations. It is this later method (assisted by CAS), that is outlined here.
Historical Context
In 1753 Euler integrated the differential equation [1] \begin{equation} \label{eq:lem_de} \frac {dx} {\sqrt{1-x^4}} \space = \space \frac {dy} {\sqrt{1-y^4}} \end{equation} to the form \begin{equation} \label{eq:lem_add} x \space = \space \frac {y\sqrt{1 - c^4} + c\sqrt{1 - y^4}} {1 + c^2 y^2} \end{equation} where $c$ is the constant of integration. Utilising the identity \begin{equation*} \prod_{\textsf{all signs}} (a \pm b \pm c) \space = \space \left(a^2 + b^2 + c^2\right)^2 \space - \space 4\left(a^2b^2 + a^2c^2 + b^2c^2\right) \end{equation*} where all signs means all 4 terms that arise from different combinations of plus and minus signs, we can convert \eqref{eq:lem_add} to a symmetric polynomial $A$ \begin{equation} \label{eq:lem_sym} A(x,y) \space = \space \left(x^2y^2c^2 + x^2 + y^2 + c^2\right)^2 \space - \space 4\cdot\left(x^2y^2 + x^2c^2 + y^2c^2\right) \space = \space 0 \end{equation} Then differential equation \eqref{eq:lem_de} can be interpreted as a scaled version of the first degree term in the Taylor expansion of $A$ \begin{equation} \frac {\partial A} {\partial x} dx \space + \space \frac {\partial A} {\partial y} dy \space = \space 0 \end{equation}
Some differential equations like \eqref{eq:lem_de} have a simple geometric interpretation. Consider the cubic curve given by \begin{equation} \label{eq:cubic} y^2 \space = \space K(x - e_1)(x - e_2)(x - e_3) \end{equation} Intersect it with a straight line passing through three points with $x$-coordinates $x_1,x_2,x_3$. Jiggle the $x$-coordinates of the first two points of intersection by infinitesimal amounts $dx_1$ and $dx_2$, then the infinitesimal movement $dx_3$, of the third point of intersection, will be constrained to satisfy the differential equation \begin{equation} \label{eq:cubic_de} \frac {dx_1} {\sqrt{(x_1-e_1)(x_1-e_2)(x_1-e_3)}} \enspace + \enspace \frac {dx_2} {\sqrt{(x_2-e_1)(x_2-e_2)(x_2-e_3)}} \enspace + \enspace \frac {dx_3} {\sqrt{(x_3-e_1)(x_3-e_2)(x_3-e_3)}} \enspace = \enspace 0 \end{equation} This type of construction was used by Abel to prove his addition theorem for algebraic integrals. Although \eqref{eq:cubic_de} is a simple algebraic identity, it's direct verification is lengthy computation.
Cubic Algebraic Integral
The condition that the three points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ lie on a straight line is given by \begin{equation} \label{eq:line} \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3 \\ \end{vmatrix} \space = \space 0 \end{equation} When these three points also lie on the cubic curve \eqref{eq:cubic} we have \begin{equation} \label{eq:cubic_add} \begin{vmatrix} 1 & x_1 & \sqrt{(x_1-e_1)(x_1-e_2)(x_1-e_3)} \\ 1 & x_2 & \sqrt{(x_2-e_1)(x_2-e_2)(x_2-e_3)} \\ 1 & x_3 & \sqrt{(x_3-e_1)(x_3-e_2)(x_3-e_3)} \\ \end{vmatrix} \space = \space 0 \end{equation} This means that \eqref{eq:cubic_add} is an algebraic integral of differential equation \eqref{eq:cubic_de}. This formula contains radicals and trivially vanishes when $x_1 = x_2$ etc. Like \eqref{eq:lem_add} it can be rationalised to a formula \begin{equation} \label{eq:delta} \Delta(x_1,x_2,x_3,e_1,e_2,e_3) \enspace = \enspace 0 \end{equation} as follows \begin{equation} \label{eq:cubic_sym} \begin{aligned} \Delta(x_1,x_2,x_3,e_1,e_2,e_3) \enspace &= \enspace \left. \prod\limits_{\textsf{all signs}} \begin{vmatrix} 1 & x_1 & \hphantom{+}\sqrt{(x_1-e_1)(x_1-e_2)(x_1-e_3)} \\ 1 & x_2 & \pm\sqrt{(x_2-e_1)(x_2-e_2)(x_2-e_3)} \\ 1 & x_3 & \pm\sqrt{(x_3-e_1)(x_3-e_2)(x_3-e_3)} \\ \end{vmatrix} \enspace \middle/ \enspace \begin{vmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \\ \end{vmatrix}^2 \right. \\\\ &= \enspace \Big[ (x_1x_2 + x_1x_3 + x_2x_3) \space - \space (e_1e_2 + e_1e_3 + e_2e_3) \Big]^2 \space - \space 4 \cdot (x_1x_2x_3 \space - \space e_1e_2e_3) \cdot \Big[ (x_1+x_2+x_3) \space - \space (e_1+e_2+e_3) \Big] \end{aligned} \end{equation} In an unexpected turn of events we find that the right-hand side of \eqref{eq:cubic_sym} is symmetric in $x \leftrightarrow e$. This implies the identity \begin{equation} \label{eq:identity} \begin{vmatrix} 1 & e_1 & e_1^2 \\ 1 & e_2 & e_2^2 \\ 1 & e_3 & e_3^2 \\ \end{vmatrix}^2 \cdot \prod_{\textsf{all signs}} \begin{vmatrix} 1 & x_1 & \hphantom{+}\sqrt{(x_1-e_1)(x_1-e_2)(x_1-e_3)} \\ 1 & x_2 & \pm\sqrt{(x_2-e_1)(x_2-e_2)(x_2-e_3)} \\ 1 & x_3 & \pm\sqrt{(x_3-e_1)(x_3-e_2)(x_3-e_3)} \\ \end{vmatrix} \enspace = \enspace \begin{vmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \\ \end{vmatrix}^2 \cdot \prod_{\textsf{all signs}} \begin{vmatrix} 1 & e_1 & \hphantom{+}\sqrt{(x_1-e_1)(x_2-e_1)(x_3-e_1)} \\ 1 & e_2 & \pm\sqrt{(x_1-e_2)(x_2-e_2)(x_3-e_2)} \\ 1 & e_3 & \pm\sqrt{(x_1-e_3)(x_2-e_3)(x_3-e_3)} \\ \end{vmatrix} \end{equation} Therefore interchanging $x \leftrightarrow e$ in \eqref{eq:cubic_add} yields another determinant style algebraic integral of \eqref{eq:cubic_de} namely \begin{equation} \label{eq:cubic_dual} \begin{vmatrix} 1 & e_1 & \sqrt{(x_1-e_1)(x_2-e_1)(x_3-e_1)} \\ 1 & e_2 & \sqrt{(x_1-e_2)(x_2-e_2)(x_3-e_2)} \\ 1 & e_3 & \sqrt{(x_1-e_3)(x_2-e_3)(x_3-e_3)} \\ \end{vmatrix} \enspace = \enspace 0 \end{equation}
Symmetric Determinant
In view of this we might try to find an algebraic determinant symmetric in $x$ and $e$ which rationalises to $\Delta(x_i,x_2,x_3,e_1,e_2,e_3)$. A reasonable guess is \begin{equation} \label{eq:cubic_trivial} \begin{vmatrix} \sqrt{x_1-e_1} & \sqrt{x_1-e_2} & \sqrt{x_1-e_3} \\ \sqrt{x_2-e_1} & \sqrt{x_2-e_2} & \sqrt{x_2-e_3} \\ \sqrt{x_3-e_1} & \sqrt{x_3-e_2} & \sqrt{x_3-e_3} \\ \end{vmatrix} \enspace = \enspace 0 \end{equation} but as the right-hand side shows \begin{equation} \label{eq:cubic_trivial_identity} \prod_{\textsf{all signs}} \begin{vmatrix} \sqrt{x_1-e_1} & \hphantom{+}\sqrt{x_1-e_2} & \hphantom{+}\sqrt{x_1-e_3} \\ \sqrt{x_2-e_1} & \pm\sqrt{x_2-e_2} & \pm\sqrt{x_2-e_3} \\ \sqrt{x_3-e_1} & \pm\sqrt{x_3-e_2} & \pm\sqrt{x_3-e_3} \\ \end{vmatrix} \enspace = \enspace 2^{8} \cdot \begin{vmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \\ \end{vmatrix}^{4} \cdot \enspace \begin{vmatrix} 1 & e_1 & e_1^2 \\ 1 & e_2 & e_2^2 \\ 1 & e_3 & e_3^2 \\ \end{vmatrix}^{4} \end{equation} this fails to produce any factor of $\Delta$.
Fully Symmetric Cubic Algebraic Integral
With the help of elliptic function theory, the $x \leftrightarrow e$ symmetric determinant and algebraic integral of \eqref{eq:cubic_de} we seek, is given by \begin{equation} \label{eq:cubic_super} \begin{vmatrix} 0 & 1 & 1 & 1 \\ 1 & \sqrt{x_1-e_1} & \sqrt{x_1-e_2} & \sqrt{x_1-e_3} \\ 1 & \sqrt{x_2-e_1} & \sqrt{x_2-e_2} & \sqrt{x_2-e_3} \\ 1 & \sqrt{x_3-e_1} & \sqrt{x_3-e_2} & \sqrt{x_3-e_3} \\ \end{vmatrix} \enspace = \enspace 0 \end{equation} If this determinant vanishes for any set of distinct $x_i$ and set of distinct $e_i$ and combination of signs on the square roots, then $\Delta(x_1,x_2,x_3,e_1,e_2,e_3)$ also vanishes. This fact is easily seen from the following algebraic identity \begin{equation} \label{eq:cubic_super_identity} \prod_{\textsf{evens}} \begin{vmatrix} 0 & 1 & 1 & 1 \\ 1 & \hphantom{+}\sqrt{x_1-e_1} & \pm\sqrt{x_1-e_2} & \pm\sqrt{x_1-e_3} \\ 1 & \pm\sqrt{x_2-e_1} & \pm\sqrt{x_2-e_2} & \pm\sqrt{x_2-e_3} \\ 1 & \pm\sqrt{x_3-e_1} & \pm\sqrt{x_3-e_2} & \pm\sqrt{x_3-e_3} \\ \end{vmatrix} \enspace = \enspace 2^{160} \cdot \begin{vmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \\ \end{vmatrix}^{16} \cdot \enspace \begin{vmatrix} 1 & e_1 & e_1^2 \\ 1 & e_2 & e_2^2 \\ 1 & e_3 & e_3^2 \\ \end{vmatrix}^{16} \cdot \enspace \Delta(x_1,x_2,x_3,e_1,e_2,e_3)^{8} \end{equation} where evens means the product includes only determinants with an even number of minus signs. A similar identity holds for odds.
Cubic Differential Equation
We can now prove \eqref{eq:cubic_de} as follows. First note that the obvious version of this differential equation is obtained by differentiating \eqref{eq:delta} \begin{equation} \label{eq:cubic_delta_de} \frac {\partial \Delta} {\partial x_1} dx_1 \enspace + \enspace \frac {\partial \Delta} {\partial x_2} dx_2 \enspace + \enspace \frac {\partial \Delta} {\partial x_3} dx_3 \enspace = \enspace 0 \end{equation} In order for \eqref{eq:cubic_delta_de} to be equivalent to \eqref{eq:cubic_de} the coefficients of the infinitesimals must be proportional. A straight forward but lengthy computation gives \begin{equation} \label{eq:cubic_de_ratio} \left(\frac {\partial \Delta} {\partial x_1} \right)^2 (x_1-e_1)(x_1-e_2)(x_1-e_3) \enspace - \enspace \left(\frac {\partial \Delta} {\partial x_2} \right)^2 (x_2-e_1)(x_2-e_2)(x_2-e_3) \enspace = \enspace (x_1 - x_2) \cdot P(x_1,x_2,x_3,e_1,e_2,e_3) \cdot \Delta(x_1,x_2,x_3,e_1,e_2,e_3) \enspace = \enspace 0 \end{equation} where $P$ is some polynomial, and from this, and two similar pairings \eqref{eq:cubic_de} follows.
Full Cubic Differential Equation
But \eqref{eq:cubic_de} is not entirely satisfactory because it is not symmetric in $x$ and $e$. However if we jiggle both the straight line and the cubic curve then the infinitesimal variations will satisfy \begin{equation} \frac {\partial \Delta} {\partial x_1} dx_1 \enspace + \enspace \frac {\partial \Delta} {\partial x_2} dx_2 \enspace + \enspace \frac {\partial \Delta} {\partial x_3} dx_3 \enspace + \enspace \frac {\partial \Delta} {\partial e_1} de_1 \enspace + \enspace \frac {\partial \Delta} {\partial e_2} de_2 \enspace + \enspace \frac {\partial \Delta} {\partial e_3} de_3 \enspace = \enspace 0 \end{equation} A similar computation to \eqref{eq:cubic_de_ratio} then gives \begin{equation} \left(\frac {\partial \Delta} {\partial x_1} \right)^2 (x_1-e_1)(x_1-e_2)(x_1-e_3) \enspace - \enspace \left(\frac {\partial \Delta} {\partial e_1} \right)^2 (x_1-e_1)(x_2-e_1)(x_3-e_1) \enspace = \enspace (x_1 - e_1) \cdot Q(x_1,x_2,x_3,e_1,e_2,e_3) \cdot \Delta(x_1,x_2,x_3,e_1,e_2,e_3) \enspace = \enspace 0 \end{equation} where $Q$ is some polynomial. This implies that the full differential equation may be written \begin{multline} \label{eq:cubic_de_full} \frac {dx_1} {\sqrt{(x_1-e_1)(x_1-e_2)(x_1-e_3)}} \enspace + \enspace \frac {dx_2} {\sqrt{(x_2-e_1)(x_2-e_2)(x_2-e_3)}} \enspace + \enspace \frac {dx_3} {\sqrt{(x_3-e_1)(x_3-e_2)(x_3-e_3)}} \enspace + \\ \frac {de_1} {\sqrt{(x_1-e_1)(x_2-e_1)(x_3-e_1)}} \enspace + \enspace \frac {de_2} {\sqrt{(x_1-e_2)(x_2-e_2)(x_3-e_2)}} \enspace + \enspace \frac {de_3} {\sqrt{(x_1-e_3)(x_2-e_3)(x_3-e_3)}} \quad = \quad 0 \qquad\qquad \end{multline} and this differential equation has obvious $x \leftrightarrow e$ symmetry.
Quartic Curve
Now consider the quartic curve given by \begin{equation} \label{eq:quartic} y^2 \space = \space K(x - e_1)(x - e_2)(x - e_3)(x - e_4) \end{equation} Intersect it with a parabola passing through four points with $x$-coordinates $x_1,x_2,x_3,x_4$. Jiggle the $x$-coordinates of the first three points of intersection by infinitesimal amounts $dx_1,dx_2,dx_3$, then the movement of the fourth point $dx_4$ when constrained to lie on the parabola passing through the first three points, will satisfy the following differential equation \begin{multline} \label{eq:quartic_de} \frac {dx_1} {\sqrt{(x_1-e_1)(x_1-e_2)(x_1-e_3)(x_1-e_4)}} \enspace + \enspace \frac {dx_2} {\sqrt{(x_2-e_1)(x_2-e_2)(x_2-e_3)(x_2-e_4)}} \enspace + \enspace \\ \frac {dx_3} {\sqrt{(x_3-e_1)(x_3-e_2)(x_3-e_3)(x_3-e_4)}} \enspace + \enspace \frac {dx_4} {\sqrt{(x_4-e_1)(x_4-e_2)(x_4-e_3)(x_4-e_4)}} \enspace = \enspace 0 \end{multline} The condition that four points lie on a parabola is \begin{equation} \label{eq:parabola} \begin{vmatrix} 1 & x_1 & x_1^2 & y_1 \\ 1 & x_2 & x_2^2 & y_2 \\ 1 & x_3 & x_3^2 & y_3 \\ 1 & x_4 & x_4^2 & y_4 \\ \end{vmatrix} \enspace = \enspace 0 \end{equation} When these four points also lie on the quartic curve \eqref{eq:quartic} we have \begin{equation} \label{eq:quartic_add} \begin{vmatrix} 1 & x_1 & x_1^2 & \sqrt{(x_1 - e_1)(x_1 - e_2)(x_1 - e_3)(x_1 - e_4)} \\ 1 & x_2 & x_2^2 & \sqrt{(x_2 - e_1)(x_2 - e_2)(x_2 - e_3)(x_2 - e_4)} \\ 1 & x_3 & x_3^2 & \sqrt{(x_3 - e_1)(x_3 - e_2)(x_3 - e_3)(x_3 - e_4)} \\ 1 & x_4 & x_4^2 & \sqrt{(x_4 - e_1)(x_4 - e_2)(x_4 - e_3)(x_4 - e_4)} \\ \end{vmatrix} \enspace = \enspace 0 \end{equation} and therefore \eqref{eq:quartic_add} is an algebraic integral of \eqref{eq:quartic_de}. As was done previously this formula can rationalised to a formula \begin{equation} \label{eq:fraktur_R} \mathfrak{R}(x_1,x_2,x_3,x_4,e_1,e_2,e_3,e_4) \enspace = \enspace 0 \end{equation} where \begin{equation} \begin{aligned} \mathfrak{R}(x_1,x_2,x_3,x_4,e_1,e_2,e_3,e_4) \enspace &= \enspace \begin{vmatrix} 1 & x_1 & x_1^2 & x_1^3 \\ 1 & x_2 & x_2^2 & x_2^3 \\ 1 & x_3 & x_3^2 & x_3^3 \\ 1 & x_4 & x_4^2 & x_4^3 \\ \end{vmatrix}^{-4} \prod_{\textsf{all signs}} \begin{vmatrix} 1 & x_1 & x_1^2 & \hphantom{+} \sqrt{(x_1 - e_1)(x_1 - e_2)(x_1 - e_3)(x_1 - e_4)} \\ 1 & x_2 & x_2^2 & \pm \sqrt{(x_2 - e_1)(x_2 - e_2)(x_2 - e_3)(x_2 - e_4)} \\ 1 & x_3 & x_3^2 & \pm \sqrt{(x_3 - e_1)(x_3 - e_2)(x_3 - e_3)(x_3 - e_4)} \\ 1 & x_4 & x_4^2 & \pm \sqrt{(x_4 - e_1)(x_4 - e_2)(x_4 - e_3)(x_4 - e_4)} \\ \end{vmatrix} \\\\ &= \enspace \left(x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4\right)^4\left(e_1 + e_2 + e_3 + e_4\right)^4 \quad + \quad \text{another 237 similar terms} \end{aligned} \end{equation}
Simplification
Using the evens/odds method, this can be simplified as follows, let \begin{equation} \begin{aligned} \mathfrak{S}(x_1,x_2,x_3,x_4,e_1,e_2,e_3,e_4) \enspace &= \enspace \frac 1 2 \cdot \begin{vmatrix} 1 & x_1 & x_1^2 & x_1^3 \\ 1 & x_2 & x_2^2 & x_2^3 \\ 1 & x_3 & x_3^2 & x_3^3 \\ 1 & x_4 & x_4^2 & x_4^3 \\ \end{vmatrix}^{-2} \cdot \space \left( \prod_{\textsf{evens}} \begin{vmatrix} 1 & x_1 & x_1^2 & \hphantom{+} \sqrt{(x_1 - e_1)(x_1 - e_2)(x_1 - e_3)(x_1 - e_4)} \\ 1 & x_2 & x_2^2 & \pm \sqrt{(x_2 - e_1)(x_2 - e_2)(x_2 - e_3)(x_2 - e_4)} \\ 1 & x_3 & x_3^2 & \pm \sqrt{(x_3 - e_1)(x_3 - e_2)(x_3 - e_3)(x_3 - e_4)} \\ 1 & x_4 & x_4^2 & \pm \sqrt{(x_4 - e_1)(x_4 - e_2)(x_4 - e_3)(x_4 - e_4)} \\ \end{vmatrix} \enspace + \enspace \prod_{\textsf{odds}} \ldots \right) \\\\ &= \enspace\left(x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4\right)^2\left(e_1 + e_2 + e_3 + e_4\right)^2 \quad + \quad \text{another 23 similar terms} \end{aligned} \end{equation} Let $R$ and $S$ be the two monic quartic polynomials with roots given by the $x_i$ and $e_i$. Then utilising the identities \begin{equation*} \prod_{\textsf{evens/odds}} (a \pm b \pm c \pm d) \space = \space \left(a^2 + b^2 + c^2 + d^2\right)^2 \space - \space 4\left(a^2b^2 + a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 + c^2d^2\right) \space \pm \space 8abcd \end{equation*} we have \begin{equation} \label{eq:quartic_simplified} \mathfrak{R}(R,S) \enspace = \enspace \mathfrak{S}(R,S)^2 \enspace - \enspace 64 \cdot \resultant(R,S) \end{equation} Formula \eqref{eq:quartic_simplified} also leads to this identity \begin{equation} \label{eq:quartic_evens} \prod_{\textsf{evens/odds}} \begin{vmatrix} 1 & x_1 & x_1^2 & \hphantom{\pm}\sqrt{(x_1 - e_1)(x_1 - e_2)(x_1 - e_3)(x_1 - e_4)} \\ 1 & x_2 & x_2^2 & \pm\sqrt{(x_2 - e_1)(x_2 - e_2)(x_2 - e_3)(x_2 - e_4)} \\ 1 & x_3 & x_3^2 & \pm\sqrt{(x_3 - e_1)(x_3 - e_2)(x_3 - e_3)(x_3 - e_4)} \\ 1 & x_4 & x_4^2 & \pm\sqrt{(x_4 - e_1)(x_4 - e_2)(x_4 - e_3)(x_4 - e_4)} \\ \end{vmatrix} \enspace = \enspace \begin{vmatrix} 1 & x_1 & x_1^2 & x_1^3 \\ 1 & x_2 & x_2^2 & x_2^3 \\ 1 & x_3 & x_3^2 & x_3^3 \\ 1 & x_4 & x_4^2 & x_4^3 \\ \end{vmatrix}^2 \left[ \mathfrak{S}(R,S) \enspace \pm \enspace 8 \cdot \sqrt{\resultant(R,S)} \right] \end{equation} Which shows how every quartic formula is effectively the product of two versions of an equivalent cubic formula. When say $x_4 = e_4$ the resultant vanishes and the two copies become identical allowing the formula to be reduced by taking the square root.
Cross Check
Cross-checking \eqref{eq:quartic_simplified} against the Euler formula \eqref{eq:lem_sym} gives \begin{equation*} \begin{aligned} \frac 1 {64} \cdot \mathfrak{R}(x,y,c,0,+1,-1,+\imath,-\imath) \enspace &= \enspace \left(x^2y^2+x^2c^2+y^2c^2-1\right)^2 \enspace - \enspace (1-x^4)(1-y^4)(1-c^4) \\\\ &= \enspace \left(x^2y^2c^2 + x^2 + y^2 + c^2\right)^2 \enspace - \enspace 4\cdot\left(x^2y^2 + x^2c^2 + y^2c^2\right) \qquad\qquad\qquad\qquad \large\checkmark \end{aligned} \end{equation*}
Connection With Classical Invariant Theory
Unlike the equations associated with the cubic curve, the quartic equations are invariant under Möbius transformations. This means $\mathfrak{R}$ and $\mathfrak{S}$ are simultaneous invariants of the two polynomials $R$ and $S$. The invariant $\mathfrak{S}$ can be concisely expressed in terms of scaled transvectants, see The invariants of 2V4 [2] \begin{equation} \label{eq:transvectant} \mathfrak{S}(R,S) \enspace = \enspace 96 \thinspace \transvectant{\transvectant{R,R}_2,\transvectant{S,S}_2}_4 \enspace - \enspace 8 \thinspace \transvectant{R,S}_4^2 \enspace - \enspace 8 \thinspace \transvectant{R,R}_4 \transvectant{S,S}_4 \end{equation} Because the terms in \eqref{eq:quartic_simplified} and \eqref{eq:transvectant} are symmetric under $R \leftrightarrow S$ this implies yet more unexpected symmetry, namely \begin{equation} \mathfrak{S}(x_1,x_2,x_3,x_4,e_1,e_2,e_3,e_4) \space = \space \mathfrak{S}(e_1,e_2,e_3,e_4,x_1,x_2,x_3,x_4)\hspace{4em} \textsf{and} \hspace{4em} \mathfrak{R}(x_1,x_2,x_3,x_4,e_1,e_2,e_3,e_4) \space = \space \mathfrak{R}(e_1,e_2,e_3,e_4,x_1,x_2,x_3,x_4) \end{equation} This symmetry implies another algebraic integral of \eqref{eq:quartic_de} is given by \begin{equation} \begin{vmatrix} 1 & e_1 & e_1^2 & \sqrt{(x_1 - e_1)(x_2 - e_1)(x_3 - e_1)(x_4 - e_1)} \\ 1 & e_2 & e_2^2 & \sqrt{(x_1 - e_2)(x_2 - e_2)(x_3 - e_2)(x_4 - e_2)} \\ 1 & e_3 & e_3^2 & \sqrt{(x_1 - e_3)(x_2 - e_3)(x_3 - e_3)(x_4 - e_3)} \\ 1 & e_4 & e_4^2 & \sqrt{(x_1 - e_4)(x_2 - e_4)(x_3 - e_4)(x_4 - e_4)} \\ \end{vmatrix} \enspace = \enspace 0 \end{equation}
Connection Between Cubic And Quartic Formulae
Using elementary row and column operations it is easy to evaluate $\mathfrak{S}$ at $x_4 = e_4$ giving \begin{equation} \mathfrak{S}(x_1,x_2,x_3,e_4,e_1,e_2,e_3,e_4) \enspace = \enspace \begin{vmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \\ \end{vmatrix}^{-2} \cdot \space (x_1 - e_4)^2(x_2 - e_4)^2(x_3 - e_4)^2 \space \cdot \prod_{\textsf{all signs}} \begin{vmatrix} 1 & x_1 & \hphantom{+} \sqrt{(x_1 - e_1)(x_1 - e_2)(x_1 - e_3)/(x_1 - e_4)} \\ 1 & x_2 & \pm \sqrt{(x_2 - e_1)(x_2 - e_2)(x_2 - e_3)/(x_2 - e_4)} \\ 1 & x_3 & \pm \sqrt{(x_3 - e_1)(x_3 - e_2)(x_3 - e_3)/(x_3 - e_4)} \\ \end{vmatrix} \end{equation} From this formula it is easy to see that \begin{equation} \begin{aligned} \lim_{t \rightarrow \infty} \frac {\mathfrak{S}(x_1,x_2,x_3,t,e_1,e_2,e_3,t)} {t^4} \enspace = \enspace \Delta(x_1,x_2,x_3,e_1,e_2,e_3) \end{aligned} \end{equation} and because the resultant vanishes in \eqref{eq:quartic_simplified} \begin{equation} \begin{aligned} \lim_{t \rightarrow \infty} \frac {\mathfrak{R}(x_1,x_2,x_3,t,e_1,e_2,e_3,t)} {t^8} \enspace = \enspace \Delta(x_1,x_2,x_3,e_1,e_2,e_3)^2 \end{aligned} \end{equation}
Intermediate Quartic Determinant
Another algebraic integral for \eqref{eq:quartic_de} is given by \begin{equation} \label{eq:quartic_sqrtx2} \begin{vmatrix} 1 & x_1 & \sqrt{(x_1-e_1)(x_1-e_2)} & \sqrt{(x_1-e_3)(x_1-e_4)} \\ 1 & x_2 & \sqrt{(x_2-e_1)(x_2-e_2)} & \sqrt{(x_2-e_3)(x_2-e_4)} \\ 1 & x_3 & \sqrt{(x_3-e_1)(x_3-e_2)} & \sqrt{(x_3-e_3)(x_3-e_4)} \\ 1 & x_4 & \sqrt{(x_4-e_1)(x_4-e_2)} & \sqrt{(x_4-e_3)(x_4-e_4)} \\ \end{vmatrix} \enspace = \enspace 0 \end{equation} This fact is confirmed by the identity \begin{equation} \label{eq:quartic_double} \prod_{\textsf{all signs}} \begin{vmatrix} 1 & x_1 & \hphantom{+}\sqrt{(x_1-e_1)(x_1-e_2)} & \hphantom{+}\sqrt{(x_1-e_3)(x_1-e_4)} \\ 1 & x_2 & \pm\sqrt{(x_2-e_1)(x_2-e_2)} & \pm\sqrt{(x_2-e_3)(x_2-e_4)} \\ 1 & x_3 & \pm\sqrt{(x_3-e_1)(x_3-e_2)} & \pm\sqrt{(x_3-e_3)(x_3-e_4)} \\ 1 & x_4 & \pm\sqrt{(x_4-e_1)(x_4-e_2)} & \pm\sqrt{(x_4-e_3)(x_4-e_4)} \\ \end{vmatrix} \enspace = \enspace 2^{32} \cdot \begin{vmatrix} 1 & x_1 & x_1^2 & x_1^3 \\ 1 & x_2 & x_2^2 & x_2^3 \\ 1 & x_3 & x_3^2 & x_3^3 \\ 1 & x_4 & x_4^2 & x_4^3 \\ \end{vmatrix}^{16} \hspace{-0.7em} \cdot (e_1-e_2)^{16}(e_3-e_4)^{16} \cdot \mathfrak{R}(x_1,x_2,x_3,x_4,e_1,e_2,e_3,e_4)^{4} \end{equation} Symmetry in this formula gives another integral for \eqref{eq:quartic_de} \begin{equation} \begin{vmatrix} 1 & e_1 & \sqrt{(x_1-e_1)(x_2-e_1)} & \sqrt{(x_3-e_1)(x_4-e_1)} \\ 1 & e_2 & \sqrt{(x_1-e_2)(x_2-e_2)} & \sqrt{(x_3-e_2)(x_4-e_2)} \\ 1 & e_3 & \sqrt{(x_1-e_3)(x_2-e_3)} & \sqrt{(x_3-e_3)(x_4-e_3)} \\ 1 & e_4 & \sqrt{(x_1-e_4)(x_2-e_4)} & \sqrt{(x_3-e_4)(x_4-e_4)} \\ \end{vmatrix} \enspace = \enspace 0 \end{equation}
Fully Symmetric Quartic Determinant
A fully symmetric algebraic integral for \eqref{eq:quartic_de} is given by \begin{equation} \label{eq:quartic_super} \begin{vmatrix} \sqrt{x_1-e_1} & \sqrt{x_1-e_2} & \sqrt{x_1-e_3} & \sqrt{x_1-e_4} \\ \sqrt{x_2-e_1} & \sqrt{x_2-e_2} & \sqrt{x_2-e_3} & \sqrt{x_2-e_4} \\ \sqrt{x_3-e_1} & \sqrt{x_3-e_2} & \sqrt{x_3-e_3} & \sqrt{x_3-e_4} \\ \sqrt{x_4-e_1} & \sqrt{x_4-e_2} & \sqrt{x_4-e_3} & \sqrt{x_4-e_4} \\ \end{vmatrix} \enspace = \enspace 0 \end{equation} If this determinant vanishes for any distinct $x_i$ and $e_i$ and combination of signs on the square roots, then $\mathfrak{R}(x_1,x_2,x_3,x_4,e_1,e_2,e_3,e_4)$ also vanishes. This fact is easily seen from the following algebraic identity \begin{equation} \label{eq:quartic_super_identity} \prod_{\textsf{all signs}} \begin{vmatrix} \sqrt{x_1-e_1} & \hphantom{+}\sqrt{x_1-e_2} & \hphantom{+}\sqrt{x_1-e_3} & \hphantom{+}\sqrt{x_1-e_4} \\ \sqrt{x_2-e_1} & \pm\sqrt{x_2-e_2} & \pm\sqrt{x_2-e_3} & \pm\sqrt{x_2-e_4} \\ \sqrt{x_3-e_1} & \pm\sqrt{x_3-e_2} & \pm\sqrt{x_3-e_3} & \pm\sqrt{x_3-e_4} \\ \sqrt{x_4-e_1} & \pm\sqrt{x_4-e_2} & \pm\sqrt{x_4-e_3} & \pm\sqrt{x_4-e_4} \\ \end{vmatrix} \enspace = \enspace 2^{640} \cdot \begin{vmatrix} 1 & x_1 & x_1^2 & x_1^3 \\ 1 & x_2 & x_2^2 & x_2^3 \\ 1 & x_3 & x_3^2 & x_3^3 \\ 1 & x_4 & x_4^2 & x_4^3 \\ \end{vmatrix}^{64} \hspace{-0.5em} \cdot \space \begin{vmatrix} 1 & e_1 & e_1^2 & e_1^3 \\ 1 & e_2 & e_2^2 & e_2^3 \\ 1 & e_3 & e_3^2 & e_3^3 \\ 1 & e_4 & e_4^2 & e_4^3 \\ \end{vmatrix}^{64} \hspace{-0.5em} \cdot \space \mathfrak{R}(x_1,x_2,x_3,x_4,e_1,e_2,e_3,e_4)^{16} \end{equation} The evens/odds method does not yield a simpler rational function here, instead we get formulae similar to \eqref{eq:quartic_evens}
Quartic Differential Equation
We can now prove \eqref{eq:quartic_de} as follows. First note that the obvious version of this differential equation is obtained by differentiating \eqref{eq:fraktur_R} \begin{equation} \label{eq:quartic_R_de} \frac {\partial \mathfrak{R}} {\partial x_1} dx_1 \enspace + \enspace \frac {\partial \mathfrak{R}} {\partial x_2} dx_2 \enspace + \enspace \frac {\partial \mathfrak{R}} {\partial x_3} dx_3 \enspace + \enspace \frac {\partial \mathfrak{R}} {\partial x_4} dx_4 \enspace = \enspace 0 \end{equation} In order for \eqref{eq:fraktur_R} to be equivalent to \eqref{eq:quartic_de} the coefficients of the infinitesimals must be proportional. A straight forward but lengthy computation gives \begin{equation} \label{eq:quartic_de_ratio} \left(\frac {\partial \mathfrak{R}} {\partial x_1} \right)^2 S(x_1) \enspace - \enspace \left(\frac {\partial \mathfrak{R}} {\partial x_2} \right)^2 S(x_2) \enspace = \enspace (x_1 - x_2) \cdot P(x_1,x_2,x_3,x_4,e_1,e_2,e_3,e_4) \cdot \mathfrak{R}(x_1,x_2,x_3,x_4,e_1,e_2,e_3,e_4) \enspace = \enspace 0 \end{equation} where $P$ is some polynomial, and from this, and five similar pairings \eqref{eq:quartic_de} follows.
Anharmonic Cubic Curve
This anharmonic cubic curve \begin{equation} \label{eq:sextic} y^3 \space = \space K (x - e_1)(x - e_2)(x - e_3) \end{equation} exhibits similar symmetries to the cubic and quartic curves above. Intersect it with a straight line passing through three points and jiggle it to get this differential equation \begin{equation} \label{eq:sextic_de} \frac {dx_1} {\sqrt[3]{(x_1-e_1)^2(x_1-e_2)^2(x_1-e_3)^2}} \enspace + \enspace \frac {dx_2} {\sqrt[3]{(x_2-e_1)^2(x_2-e_2)^2(x_2-e_3)^2}} \enspace + \enspace \frac {dx_3} {\sqrt[3]{(x_3-e_1)^2(x_3-e_2)^2(x_3-e_3)^2}} \enspace = \enspace 0 \end{equation} The condition that three points lie on a straight line is \eqref{eq:line}. When these three points also lie on the anharmonic cubic curve \eqref{eq:sextic} we have \begin{equation} \label{eq:sextic_add} \begin{vmatrix} 1 & x_1 & \sqrt[3]{(x_1 - e_1)(x_1 - e_2)(x_1 - e_3)} \\ 1 & x_2 & \sqrt[3]{(x_2 - e_1)(x_2 - e_2)(x_2 - e_3)} \\ 1 & x_3 & \sqrt[3]{(x_3 - e_1)(x_3 - e_2)(x_3 - e_3)} \\ \end{vmatrix} \enspace = \enspace 0 \end{equation} and therefore \eqref{eq:sextic_add} is an algebraic integral of \eqref{eq:sextic_de}. As previously this formula can be rationalised to \begin{equation*} \mathfrak{T}(x_1,x_2,x_3,e_1,e_2,e_3) \enspace = \enspace 0 \end{equation*} where \begin{equation} \label{eq:sextic_sym} \begin{aligned} \mathfrak{T}(x_1,x_2,x_3,e_1,e_2,e_3) \enspace &= \enspace \begin{vmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \\ \end{vmatrix}^{-3} \cdot \prod_{0 \le i,j \le 2} \begin{vmatrix} 1 & x_1 & \hphantom{\zeta^0} \sqrt[3]{(x_1 - e_1)(x_1 - e_2)(x_1 - e_3)} \\ 1 & x_2 & \zeta^i \sqrt[3]{(x_2 - e_1)(x_2 - e_2)(x_2 - e_3)} \\ 1 & x_3 & \zeta^j \sqrt[3]{(x_3 - e_1)(x_3 - e_2)(x_3 - e_3)} \\ \end{vmatrix} \\\\ &= \enspace (x_1x_2+x_1x_3+x_2x_3)^3(e_1+e_2+e_3)^3 \quad + \quad \text{another 33 similar terms} \end{aligned} \end{equation} and $\zeta = e^{2\pi\imath \over 3}$ is a primitive third root of unity.
Simplification
Grouping the determinants according to the number of cube roots of unity they contain, we have \begin{equation} \label{eq:sextic_factor} \begin{aligned} \mathfrak{U}(x_1,x_2,x_3,e_1,e_2,e_3) \enspace &= \enspace \frac 1 3 \cdot \begin{vmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \\ \end{vmatrix}^{-1} \cdot \sum_{0 \le n \le 2} \left( \prod_{\substack{0 \le i,j \le 2 \\ i + j \equiv n \mod 3}} \begin{vmatrix} 1 & x_1 & \hphantom{\zeta^0} \sqrt[3]{(x_1 - e_1)(x_1 - e_2)(x_1 - e_3)} \\ 1 & x_2 & \zeta^i \sqrt[3]{(x_2 - e_1)(x_2 - e_2)(x_2 - e_3)} \\ 1 & x_3 & \zeta^j \sqrt[3]{(x_3 - e_1)(x_3 - e_2)(x_3 - e_3)} \\ \end{vmatrix} \right) \\\\ &= \enspace (x_1x_2+x_1x_3+x_2x_3)(e_1+e_2+e_3) \quad + \quad \text{another 3 similar terms} \end{aligned} \end{equation} Let $R$ and $S$ be the monic cubic polynomials with roots $x_i$ and $e_i$. Then utilising the identities \begin{aligned} \prod_{0 \le i,j \le 2} \left(a + \zeta^i b + \zeta^j c\right) \enspace &= \enspace \left(a^3 + b^3 + c^3\right)^3 \enspace - \enspace 27 \cdot a^3b^3c^3 \\\\ \sum_{0 \le n \le 2} \space \prod_{\substack{0 \le i,j \le 2 \\ i+j \equiv n \mod 3}} \hspace{-1em} \left(a + \zeta^i b + \zeta^j c\right) \enspace &= \enspace 3 \cdot \left(a^3 + b^3 + c^3\right) \end{aligned} we have \begin{equation} \mathfrak{T}(R,S) \enspace = \enspace \mathfrak{U}(R,S)^3 \enspace + \enspace 27 \cdot \resultant(R,S) \end{equation}
Transvectant Formulae
These equations are also invariant under Mobius transformations. Therefore $\mathfrak{T}$ and $\mathfrak{U}$ can be expressed in terms of scaled transvectants, see The invariants of 2V3 [2], as follows \begin{aligned} \mathfrak{T}(R,S) \enspace &= \enspace 3^6 \cdot \transvectant{\transvectant{R,\transvectant{R,S}_2}_2,\transvectant{S,\transvectant{S,R}_2}_2}_1 \\\\ \mathfrak{U}(R,S) \enspace &= \enspace -3 \cdot \transvectant{R,S}_3 \\\\ \resultant(R,S) \enspace &= \enspace \transvectant{R,S}_3^3 \enspace + \enspace 27 \cdot \transvectant{\transvectant{R,\transvectant{R,S}_2}_2,\transvectant{S,\transvectant{S,R}_2}_2}_1 \end{aligned} The transvectants on the right-hand side are skew-symmetric in $R$ and $S$ therefore $\mathfrak{T}(R,S)=-\mathfrak{T}(S,R)$ and another algebraic integral of \eqref{eq:sextic_de} is \begin{equation} \label{eq:sextic_add_xe} \begin{vmatrix} 1 & e_1 & \sqrt[3]{(x_1 - e_1)(x_2 - e_1)(x_3 - e_1)} \\ 1 & e_2 & \sqrt[3]{(x_1 - e_2)(x_2 - e_2)(x_3 - e_2)} \\ 1 & e_3 & \sqrt[3]{(x_1 - e_3)(x_2 - e_3)(x_3 - e_3)} \\ \end{vmatrix} \enspace = \enspace 0 \end{equation}
Fully Symmetric Anharmonic Determinant
A fully symmetric algebraic integral of \eqref{eq:sextic_de} is given by \begin{equation} \label{eq:sextic_sym_xe} \begin{vmatrix} \sqrt[3]{(x_1 - e_1)} & \sqrt[3]{(x_1 - e_2)} & \sqrt[3]{(x_1 - e_3)} \\ \sqrt[3]{(x_2 - e_1)} & \sqrt[3]{(x_2 - e_2)} & \sqrt[3]{(x_2 - e_3)} \\ \sqrt[3]{(x_3 - e_1)} & \sqrt[3]{(x_3 - e_2)} & \sqrt[3]{(x_3 - e_3)} \\ \end{vmatrix} \enspace = \enspace 0 \end{equation} This follows immediately from the identity \begin{equation} \label{eq:sextic_full_sym} \prod_{0 \le i,j,k,l \le 2} \begin{vmatrix} \sqrt[3]{(x_1 - e_1)} & \hphantom{\zeta^0}\sqrt[3]{(x_1 - e_2)} & \hphantom{\zeta^0}\sqrt[3]{(x_1 - e_3)} \\ \sqrt[3]{(x_2 - e_1)} & \zeta^i\sqrt[3]{(x_2 - e_2)} & \zeta^k\sqrt[3]{(x_2 - e_3)} \\ \sqrt[3]{(x_3 - e_1)} & \zeta^j\sqrt[3]{(x_3 - e_2)} & \zeta^l\sqrt[3]{(x_3 - e_3)} \\ \end{vmatrix} \enspace = \enspace 3^{27} \cdot \begin{vmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \\ \end{vmatrix}^9 \cdot \begin{vmatrix} 1 & e_1 & e_1^2 \\ 1 & e_2 & e_2^2 \\ 1 & e_3 & e_3^2 \\ \end{vmatrix}^9 \cdot \mathfrak{T}(x_1,x_2,x_3,e_1,e_2,e_3)^3 \end{equation}
Dual Quartic
Swap the roles of the $x$-coordinates so that $x_1,x_2,x_3,x_4$ are the $x$-intercepts of the another quartic curve. And $e_1,e_2,e_3,e_4$ are the $x$-coordinates of it's intersection with some unknown parabola. Then scale the second quartic curve so that the $y$ coordinate corresponding to $e_1$ lies on the original parabola. Remarkably the $y$-coordinates all three other points will then also lie on that original parabola.
[1] A Brief History of Elliptic Integral Addition Theorems Rose-Hulman Undergraduate Mathematics Journal: Vol. 10 : Iss. 2, Article 2.
[2] Invariants of binary forms. Doctoral Dissertation: Philosophisch-Naturwissenschaftlichen Fakultät der Universität Basel