Historical Context

In 1753, from geometric considerations on the Lemniscate of Bernoulli, Euler[1] obtained the differential equation

which he then integrated[2] to the form

where $z$ is the constant of integration. By squaring twice and dividing out trivial factors this expression can be rationalised to the polynomial

Utilising a geometric construction similar to that employed by Abel to prove his addition theorem for algebraic integrals, consider the cubic curve given by

Intersect it with a straight line passing through three points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$. 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

When written solely in terms of $x_1,x_2,x_3$ cubic_de_y becomes

Compare this differential equation with differential equation lem_de. Although cubic_de is a simple algebraic identity, it's direct verification, without the use of elliptic function theory, is a lengthy algebraic 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

When these three points also lie on the cubic curve cubic we have

This means that cubic_add is an algebraic integral of differential equation cubic_de. This formula contains radicals and trivially vanishes when $x_1 = x_2$ etc. Like lem_add it can be rationalised to a formula

as follows

where all signs means all four possible combinations of the $\pm$ signs.

In an unexpected turn of events we find that the right-hand side of cubic_sym is symmetric in $x \leftrightarrow e$. This implies the identity

Therefore interchanging $x \leftrightarrow e$ in cubic_add yields another determinant style algebraic integral of cubic_de namely

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

but as the right-hand side shows

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 cubic_de we seek, is given by

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

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 cubic_de as follows. First note that the obvious version of this differential equation is obtained by differentiating delta

In order for cubic_delta_de to be equivalent to cubic_de the coefficients of the infinitesimals must be proportional. A straight forward but lengthy computation gives

where $P$ is some polynomial, and from this, and two similar pairings cubic_de follows.

Quartic Curve

Now consider the quartic curve given by

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

When these four points also lie on the quartic curve quartic we have

and therefore quartic_add is an algebraic integral of quartic_de. As was done previously this formula can rationalised to a formula

where

Simplification

Using the evens/odds method, this can be simplified as follows, let

Let $R$ and $S$ be the two monic quartic polynomials with roots given by the $x_i$ and $e_i$.

Then utilising the two identities

with

we have

Multiplying the two variants of quartic_evens_odds together then gives this identity

Cross Check

Cross-checking quartic_simplified against the Euler formula lem_sym gives

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 [3]

Because the terms in quartic_simplified and transvectant are symmetric under $R \leftrightarrow S$ this implies yet more unexpected symmetry, namely

and

This symmetry implies another algebraic integral of quartic_de is given by

Side Observation

By examining product quartic_de_sym we see that if the quartic $R$ is a perfect square then the simultaneous invariant $\mathfrak{R}(R,S)$ vanishes. Geometrically this is the case when the intersecting parabola degenerates into a pair of vertical straight lines.

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

From this formula it is easy to see that

and because the resultant vanishes in quartic_simplified

Intermediate Quartic Determinant

Another algebraic integral for quartic_de is given by

This fact is confirmed by the identity

Symmetry in this formula gives another integral for quartic_de

Fully Symmetric Quartic Determinant

A fully symmetric algebraic integral for quartic_de is given by

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

The evens/odds method does not yield a simpler rational function here, instead we get formulae similar to quartic_evens_odds

Quartic Differential Equation

We can now prove quartic_de as follows. First note that the obvious version of this differential equation is obtained by differentiating fraktur_R

In order for fraktur_R to be equivalent to quartic_de the coefficients of the infinitesimals must be proportional. A straight forward but lengthy computation gives

where $P$ is some polynomial, and from this, and five similar pairings quartic_de follows.

Anharmonic Cubic Curve

This anharmonic cubic curve

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

The condition that three points lie on a straight line is line. When these three points also lie on the anharmonic cubic curve sextic we have

and therefore sextic_add is an algebraic integral of sextic_de. As previously this formula can be rationalised to

where

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

Let $R$ and $S$ be the monic cubic polynomials with roots $x_i$ and $e_i$. Then utilising the three identities

with

and multiplying the three variants together, we have

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 [3], as follows

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 sextic_de is

Fully Symmetric Anharmonic Determinant

A fully symmetric algebraic integral of sextic_de is given by

This follows immediately from the identity

Cubic Symmetric Differential Equation

If instead of the previous arrangement, we hold the $x_1,x_2,x_3$ fixed and jiggle the intercepts $e_1,e_2,e_3$ of the cubic curve, constrained so that the points of intersections continue to lie on a straight line, then these variables will still satisfy cubic_add and because of the unexpected symmetry, the infinitesimal variations will satisfy the differential equation

Full Cubic Differential Equation

If we jiggle both the straight line and the cubic curve then the infinitesimal variations will satisfy

A similar computation to cubic_de_ratio then gives

where $Q$ is some polynomial. This implies that the full differential equation may be written

and this differential equation has obvious $x \leftrightarrow e$ symmetry.

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.

n-Quadratic Curves

This $n$-quadratic curve

also exhibits similar symmetries to the cubic and quartic curves above. Intersect it with a horizontal line passing through two points then rationalise to get

where $\zeta$ is a primitive $n$-th root of unity.

Including this example gives four instances where $x \leftrightarrow e$ symmetry or skew-symmetry holds. However this does not generalise in any simple way to quintic and higher curves. Carrying out the computation for the quintic curve (using exact algebraic number computations with CAS) shows it does not have this simple symmetry ie. the $n=5$ analogue of identity does not hold.

[1] Euler, Leonhard (1761) Observationes de comparatione arcuum curvarum irrectificibilium Euler Archive - All Works. 252.

[2] Jose Barrios (2009) A Brief History of Elliptic Integral Addition Theorems Rose-Hulman Undergraduate Mathematics Journal: Vol. 10 : Iss. 2, Article 2.

[3] M. Popoviciu Draisma (2014) Invariants of binary forms. Doctoral Dissertation: Philosophisch-Naturwissenschaftlichen Fakultät der Universität Basel