Homogeneous diophantine equation $x^3+2y^3+6xyz=3z^3$

Question

Is it known if there are infinitely (non-proportional) many integer solutions to $x^3+2y^3+6xyz=3z^3$ ?

Motivation : if true, this would provide an alternative solution to that recent MSE question, by putting $a=\frac{3z^2}{xy},b=-\frac{x^2}{yz},c=-\frac{2y^2}{xz}$.

Answer 1

It's a homogeneous equation, so if $(x,y,z)$ is a solution then $(kx,ky,kz)$ is also a solution:

\begin{eqnarray*} (kx)^3+2(ky)^3-3(kz)^3 &=& 0 \\ \\ k^3x^3 + 2k^3y^3 - 3k^3z^3 &=& 0 \\ \\ k^3(x^3 + 2y^3-3y^3) &=& 0 \end{eqnarray*}

If it has one, non-zero solution then it will have infinitely many solutions.

Clearly $(x,y,z)=(1,1,1)$ is a solution, and so $(x,y,z)=(k,k,k)$ are all solutions.

Answer 2

It seems to be possible to use the fact that $x^3+2y^3+6xyz = 3z^3$ is an elliptic curve with Weierstrass form $y^2 + 6 x y + 16 y = x^{3} - 105 x^{2}$ and that it has rank $1$ to produce integer solutions. I used the following Sage script to do so experimentally.

R.<x, y, z> = QQ[]
eq = x^3+2*y^3+6*x*y*z-3*z^3
P = [1,1,-1]

E = EllipticCurve_from_cubic(eq, P, morphism=False)
f = EllipticCurve_from_cubic(eq, P, morphism=True)

G = E.gens()[0]
for n in [2..10]: # adjust for more solutions
    P = f.inverse()(n*G)
    a = P[0]
    k = a.denominator()
    a = a * k
    b = P[1] * k
    c = P[2] * k
    if a in ZZ and b in ZZ and c in ZZ:
        print a,b,c, a^3+2*b^3+6*a*b*c-3*c^3

I am not sure if it can be proven or not, that this will produce infinitely many non-proportional integer solutions, but atleast it seems to do so experimentally.

Answer 3

The question asked is

Is it known if there are infinitely (non-proportional) many integer solutions to $x^3+2y^3+6xyz=3z^3$ ?

This is a special case of the general equation $\,Ax^3 + By^3 + Cz^3 + 3Dxyz = 0\,$ studied by Edouard Lucas on page $15$ in the document at URL

Recherches sur L'Analyse Indeterminee et L'Arithmetique de Diophante.

He first gives a formula for getting a new points from a given point:

$$X = x(By^3 - Cz^3),\\ Y = y(Cz^3 - Ax^3),\\ Z = x(Ax^3 - By^3).$$

Then gives as an example the special case $x^3 + 2y^3 + 3z^3 = 6xyz$.

Starting from $(1,1,1)$ he produces three other integer solutions: $(1,-2,1),\:$ $(19,4,-17),\:$ $(282473,-86392,-114427)\:$ using his duplication formula. This already answers the question asked (with $z$ replaced by $-z$).

I use generalized Somos-4 sequences to produce all solutions. Define four integer sequences

$$ u_0 = 0,\, u_1 = 1,\, u_2 = 2,\, u_3 = 105,\,u_4 = 1292,\, u_n = (-1)^n u_{-n},\\ u_n = (-4u_{n-1}u_{n-3} + 105u_{n-2}u_{n-2})/u_{n-4} \quad \text{ for }\quad n>4,\\ x_n = (-1)^n (u_{n-1} u_n u_{n+2} + 2u_{n-1} u_{n+1}^2 + 19u_n^2 u_{n+1})/2, \\ y_n = (-1)^n (- u_{n-1} u_n u_{n+2} + u_{n-1} u_{n+1}^2 + 2u_n^2 u_{n+1}), \\ z_n = (-1)^n (- u_{n-1} u_n u_{n+2} - 2u_{n-1} u_{n+1}^2 + 17u_n^2 u_{n+1})/2. $$

Then $ x_n^3 + 2y_n^3 + 6x_n y_n z_n = 3z_n^3 $ for all integer $n$. Hence there are infinitely many solutions.

This is small table of values of the four sequences: $$\begin{array}{|c|c|c|c|c|} \hline n & x & y & z & u \\ \hline -3 & 282473 & -86392 & 114427 & 105 \\ \hline -2 & 143 & 113 & -71 & -2 \\ \hline -1 & 1 & -2 & -1 & 1 \\ \hline 0 & 1 & 1 & -1 & 0 \\ \hline 1 & -19 & -4 & -17 & 1 \\ \hline 2 & 16307 & 9281 & -8747 & 2 \\ \hline 3 & -259124723 & 209103562 & 2727323 & 105 \\ \hline \end{array}$$