Solving this $4$ variable diophantine equation

Question

Is there a way to solve this Diophantine equation in $a,b,c,d$? $$19a^3-33a^2b+3a^2c+30a^2d+21ab^2+24abc-12abd-15ac^2-54acd-30ad^2+ $$ $$2b^3-12b^2c-6b^2d+42bc^2+108bcd+60bd^2-7c^3-51c^2d-99cd^2-56d^3=0$$

Wolfram Alpha unfortunately cannot understand my input whenever I input this equation, and I know no strategy to solve these kinds of equations. So basically, I'm stuck...

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I want just all possible values for $a,b,c,d$ just like what Wolfram Alpha does.

Note: $a\neq b\neq c\neq d\neq 0$

Answer 1

Your equation is homogeneous of degree $3$, so if $(a,b,c,d)$ is a solution so is $(ta,tb,tc,td)$ for any $t$. Thus it makes sense to look for primitive solutions, which are solutions with greatest common divisor $1$. I found some $288$ primitive solutions with $a,b,c \in [-20\ldots -1, 1\ldots 20]$ and $d \in [1 \ldots 20]$. I don't see an obvious pattern. Here are some of those solutions:

$$ \matrix{ a & b & c & d\cr -4 & -1 & -8 & 1\cr -4 & 1 & -8 & 3\cr -3 & 1 & -7 & 3\cr -3 & 1 & 3 & 1\cr -3 & 5 & -1 & 4\cr -2 & 1 & -6 & 3\cr -2 & 1 & -4 & 2\cr -2 & 2 & -4 & 3\cr -2 & 3 & -4 & 4\cr -2 & 3 & 2 & 1\cr -2 & 4 & -4 & 5\cr -2 & 5 & -4 & 6\cr -1 & -1 & -5 & 4\cr -1 & 1 & -5 & 3\cr -1 & 2 & -3 & 3\cr -1 & 5 & 1 & 1\cr 1 & -2 & -1 & 2\cr 1 & -1 & -5 & 2\cr 1 & 1 & -3 & 3\cr 1 & 2 & -7 & 6\cr 2 & 1 & -2 & 3\cr 2 & 2 & 4 & 1\cr 2 & 3 & 4 & 2\cr 2 & 4 & 4 & 3\cr 2 & 5 & 4 & 4\cr 2 & 6 & 4 & 5\cr 3 & -4 & -7 & 1\cr 3 & 1 & -1 & 3\cr 3 & 2 & -5 & 6\cr 3 & 4 & -3 & 4\cr 4 & 3 & 8 & 1\cr 5 & 1 & 1 & 3\cr 5 & 2 & -3 & 6\cr 6 & 1 & 2 & 3\cr 7 & 1 & 3 & 3\cr 7 & 2 & -1 & 6\cr 8 & 1 & 4 & 3\cr }$$

Did those coefficients come from somewhere in particular or are they just arbitrary?

Answer 2

As the OP pointed out in the comments that the equation is equivalent to,

$$(a + 2c - 2b + 3d)^3 + (2a - b + 2c + 4d)^3 - (a + b + 2c + 2d)^3 - (3a - 2b+ c + 3d)^3 = 0$$

then by equating terms, its complete solution is,

$$\begin{aligned} a= -2 x_1 - 7 x_2 - 5 x_3 - 8 x_4\\ b=-6 x_1 + 5 x_2 - 2 x_3 + 2 x_4\\ c=9 x_1 - 14 x_2 - 10 x_3 - 3 x_4\\ d=-5 x_1 + 15 x_2 + 7 x_3 + 6 x_4 \end{aligned}$$

where,

$$x_1^3+x_2^3+x_3^3+x_4^3 = 0$$

The complete rational solution by Euler is well-known.