1

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...

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

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

Will Jagy
  • 139,541
Frank
  • 5,984

2 Answers2

3

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.

Robert Israel
  • 448,999
Tito Piezas III
  • 54,565
  • You wouldn't mind if you proved your formula: $n^3+(3n^2+2n+1)^3+(3n^3+3n^2+2n)^3=(3n^3+3n^2+2n+1)^3$? I read that in a pdf document you wrote and I'm interested to see your proof... – Frank Sep 29 '16 at 20:42
  • I'm wondering where did you get the starting grouped-equivalent from: I do not see any in the post, nor I got any reply to my comment asking exactly about that. Do you have a priviliged channel? or is it my browser's problem? – G Cab Sep 29 '16 at 21:02
  • @GCab No, I don't think the equation can be factored into your form. – Frank Sep 29 '16 at 21:32
  • @Frank: That formula in $n$? With elementary identities, the most direct way is just to expand and compare both sides. – Tito Piezas III Sep 30 '16 at 01:40
  • @GCab: The OP remarked about the $x_1^3+x_2^3+x_3^3+x_4^3=0$ form in this comment. – Tito Piezas III Sep 30 '16 at 01:44
2

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?

Robert Israel
  • 448,999
  • Robert, could it be the determinant of a 3 by 3 matrix with entries linear forms in $a,b,c,d?$ Not sure – Will Jagy Sep 29 '16 at 00:30
  • I don't know. How would one check that? – Robert Israel Sep 29 '16 at 00:35
  • Good point; it is just a possible source, the OP has not said. I do wonder whether we can join two integer solutions by a straight line and find another. Just musing. – Will Jagy Sep 29 '16 at 00:37
  • Hey; to answer your question, yes. The coefficients do come from somewhere! – Frank Sep 29 '16 at 00:40
  • Aha! In some cases you can join two solutions by a straight line and get solutions, in others you can't. For example, joining solutions $(-4,-1,-8,1)$ and $(-4,1,-8,3)$ you get a one-parameter family of solutions $(-4, 1-2t, -8, 3-2t)$. Another such family is $(2-4t, 3+t, 4-8t, 2+3t)$ from joining $(-2, 4, -4, 5)$ and $(2, 3, 4, 2)$. – Robert Israel Sep 29 '16 at 00:53
  • Robert, thanks for checking. – Will Jagy Sep 29 '16 at 00:57
  • In many cases, it seems that a two-dimensional cross-section of this cubic variety is an elliptic curve. The rational points on this elliptic curve form a group (and suitable multiples of them are then integer solutions). – Robert Israel Sep 29 '16 at 01:32
  • 2
    @RobertIsrael: The substitution $a=u+v,;c=-u+v$ reduces the OP's cubic equation to just a quadratic in $v$. Its discriminant is a quartic polynomial to be made a square, hence an elliptic curve is involved as you pointed out. – Tito Piezas III Sep 29 '16 at 01:40
  • I do not understand. Why subscribe to these figures? Write parameterization solutions! – individ Sep 29 '16 at 05:58
  • @individ OK, be my guest: parametrize all the solutions. – Robert Israel Sep 29 '16 at 07:06
  • Okay, @TitoPiezasIII Does this help: $$(a+2c-2b+3d)^3+(2a-b+2c+4d)^3=(a+b+2c+2d)^3+(3a-2b+c+3d)^3$$Using other random Diophantine equations, I have found one: $$16^3+2^3=15^3+9^3=4101$$ and using another one, I got: $$32^3+4^3=30^3+18^3=32832$$ – Frank Sep 29 '16 at 12:08
  • @RobertIsrael: With this update by the OP, it seems the equation does have a complete solution (by Euler). See my answer below. Which is strange since I thought the elliptic curve would allow infinitely many parameterizations. – Tito Piezas III Sep 29 '16 at 12:38