This week I wondered about diophantine problems that involve the volume of certain cubes and frustums, see the Wikipedia Frustum. I wondered if each one of these problems have infinitely many solutions, I would like to know if it is possible to determine if some of these have infinitely many solutions. I don't know if these problems are easily to get or if these are in the literature.

Problem 1. For integers $1<a<b$ and integers $1\leq h$ and $1\leq c$ we consider the diophantine equation $$3(a^3+c^3)=h(a^2+ab+b^2).\tag{1}$$

Example 1. One has that $31^3+7^3=30134=\frac{26}{3}(31^2+31\cdot 37+37^2)$.

Problem 2. For integers $2\leq x< y< z < v $ and integers $1\leq a < b$ and integer $h\geq 1$ we consider the diophantine equation $$3(4x^3+3y^3+2z^3+v^3)=h(a^2+ab+b^2).\tag{2}$$

Example 2. One has that the volume of four cubes of side $x=2$, added to the volume of three cubes of side $y=3$, added to the volume of two cubes $z=4$ and a cube of side $v=5$ is equals to the volume of a square frustum of basis $a=12$ and $b=15$ with height $h=2$, that is $$3(4\cdot 2^3+3\cdot 3^3+2\cdot 4^3+5^3)=1098=2(12^2+12\cdot 15+15^2).$$

Question. Do these problems, Problem 1 or Problem 2 (only is required the approach or reasoning for one of these problems), have infinite solutions? Many thanks.

I've written a small program in Pari/GP for each of these equations, showing more solutions. Each one of these codes are a line of codification that you can run in the web Sage Cell Serverchoosing as Language GP, this

for(a=2, 40, for(b=1, 40, for(c=1, 40, for(h=1, 40, if(a<b&&3*(a^3+c^3)==h*(a^2+a*b+b^2),print(c))))))

and for the second problem you get the output after two minutes

for(a=1, 15, for(b=1, 15, for(z=1,15,for(v=1,15, for(h=1, 15, for(x=2, 15, for(y=1, 15, if(a<b&&x<y&&y<z&&z<v&&3*(4*x^3+3*y^3+2*z^3+v^3)==h*(a^2+a*b+b^2),print(v)))))))))

  • $\begingroup$ All, I don't know if these problems are in the literature: diophantine problems about the volume of cubes and right frustums. To me it seems nice. Feel free to add feedback, many thanks. $\endgroup$ – user142929 Dec 5 '19 at 21:06

The Problem 1 you specify defines a cubic surface in $\mathbb{P}^{3}$, and there is a lot known about the rational points on such a surface. (For example, Chapter 2 of the book "Rational and nearly rational varieties" by Kollár, Smith and Corti is entirely about cubic surfaces.) A smooth cubic surface can fail to have any rational points at all, but if it has any rational points, the cubic surface must at least be unirational (as shown in 2000 by Janós Kollár, though for the case of the field $\mathbb{Q}$ the result may be older).

The cubic surface in Problem 1 is not smooth: it has a node at $(a : b : c : h) = (0 : 0 : 0 : 1)$, but (as one can determine nowadays with Magma), it is rational, with parametrization \begin{align*} a &= u(u^{2} + uv + v^{2})\\ b &= v(u^{2} + uv + v^{2})\\ c &= w(u^{2} + uv + v^{2})\\ h &= 3u^{3} + 3w^{3}. \end{align*} It is clear that if one takes $u$, $v$ and $w$ to be positive integers with $v > u$, then one obtains a positive integer solution satisfying your desired constraints and that this gives infinitely many positive solutions in $\mathbb{P}^{3}$.

For Problem 2, you have even more variables, and one should be able to easily treat that case using similar techniques. (One could reduce to a cubic surface by first choosing $y$, $z$ and $w$ to be linear forms in $a$, $b$, $h$ and $x$.)

  • $\begingroup$ Many thanks I'm going to refresh the definitions, and also when I can I'm going to read the statements from your references, it's incredible the power of algebraic geometry and the software. Also your knowledges! Many thanks, I'm going to wait if there is more feedback and to refresh the definitions, but your answer is very good. $\endgroup$ – user142929 Dec 6 '19 at 1:58
  • $\begingroup$ I think that your answer is excellent and the best possible, therefore I will accept it, thank you very much again. $\endgroup$ – user142929 Dec 30 '19 at 19:10

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