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The Diophantine equation $x^3+y^3=z^3+w^3$ and the Ramanujan number $1729$.

Can you please not only tell me, but also show me how to find solutions to such a Diophantine equation, for example, through elliptic curves or maybe through Eisenstein numbers. Please find a couple of solutions in your answer so I can see how it's done. As for the number $1729$. How to find and show that the minimum solution to this equation will be $(10;9;12;1)$. I would be incredibly grateful for the answer!

Angelo
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Timofey Plastinin
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  • To my knowledge taxicab numbers were all determined by brute force (either manually for ta(2) or by computers above that). – zwim Mar 12 '23 at 21:43
  • I have information that this can be done by adding points on an elliptic curve, but I don't know how to do it – Timofey Plastinin Mar 12 '23 at 22:07
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    See https://oeis.org/A001235 and the links & references given there. – Gerry Myerson Mar 12 '23 at 23:11
  • There is a parameterization of such solutions. I managed to get a quadratic one. The more interesting question is not about solutions, but what forms the parameterization of such solutions takes. https://math.stackexchange.com/questions/4591666/solving-cubic-systems-of-diophantine-equations/4595633#4595633 – individ Mar 13 '23 at 05:08
  • You may want to check this link: https://mathworld.wolfram.com/DiophantineEquation3rdPowers.html – user25406 Mar 13 '23 at 14:44
  • Any thoughts about the comments, Gang? – Gerry Myerson Mar 14 '23 at 03:19
  • Thank you very much for your answers, but I needed something specific. In some articles, the steps in finding are omitted, in the answers to stackexchange there is no finding of specific solutions and doesn't say about the number 1729 – Timofey Plastinin Mar 14 '23 at 12:26
  • See Choudhry A (1998) On Equal Sums of Cubes Theorem 2 – Adam Bailey Mar 14 '23 at 18:49
  • Did you follow-up on the link I gave, Gang, and the links and references given there? Also, if you want to be sure I see a comment intended for me, you have to put @Gerry in it. – Gerry Myerson Mar 15 '23 at 12:54
  • I looked at all the links, @Gerry, but I didn't find anything specific. – Timofey Plastinin Mar 17 '23 at 04:39
  • I don't know what you mean by "specific". The Recreomath link at the oeis page gives formulas that give an infinity of solutions. The Ono paper seems to be very much what you want. There's some advanced math there, but that's unavoidable (and anyway you said you'd be happy with a solution using elliptic curves. The Mathworld 3rd powers page looks helpful. Finding the minimal solution, as you've been told, is a simple matter of testing each number $1,2,3,\dots$ to see whether it's a sum of two cubes in two ways, and stopping when you reach $1729$. – Gerry Myerson Mar 17 '23 at 06:12

2 Answers2

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I think the smallest number for $(x^3+y^3=w^3+z^3)$ for positive numbers can be found by direct checking (or from Ramanujan view) but the following Diophantine equation can be solved using elementary methods. Define $x+y=a$, $b=x-y$, $c=z+w$ and $d=z-w$, then you have the following equation:

\begin{equation} a^3+3ab^2=c^3+3cd^2,~~~~(1) \end{equation} which can be transformed to

\begin{equation} \frac{a}{c}=\frac{c^2+3d^2}{a^2+3b^2}\to\alpha=\frac{1+3\beta^2}{\alpha^2+3\gamma^2}~~~(2) \end{equation} where we have defined $\alpha=\frac{a}{c}$, $\beta=\frac{d}{c}$ and $\gamma=\frac{b}{c}$. The original equation is a homogeneous equation, so the general solutions can be found, if we can solve the $\alpha^3+3\alpha\gamma^2=1+3\beta^2$ which is the equation (1) in terms of rational numbers $\alpha,\beta,\gamma$.

Let look at the equation (1) in the field $m+\sqrt{-3}n$ where $m,n$ are rational numbers. The equation will be simplified as follows :

\begin{equation} \alpha=\frac{1+3\beta^2}{\alpha^2+3\gamma^2}=\frac{\mid1+\sqrt{-3}\beta\mid^2}{\mid\alpha+\sqrt{-3}\gamma\mid^2}=\mid\frac{1+\sqrt{-3}\beta}{\alpha+\sqrt{-3}\gamma}\mid^2\equiv\mid A+\sqrt{-3}B\mid^2=A^2+3B^2, \end{equation}

where $\mid m+\sqrt{-3}n\mid=\sqrt{m^2+3n^2}$ is the norm of the field, so the solution of $\alpha$ in terms of $A,B$ which are rational numbers is $A^2+3B^2$. We can find $\beta$ and $\gamma$ by solving the following equations :

\begin{equation} A+\sqrt{-3}B=\frac{1+\sqrt{-3}\beta}{\alpha+\sqrt{-3}\gamma}, \end{equation} where the solutions are \begin{equation} \alpha A-3\gamma B=1\to\gamma=\frac{\alpha A-1}{3B}=\frac{A^3+3B^2A-1}{3B}, \end{equation} and \begin{equation} \beta=\alpha B+\gamma A=(A^2+3B^2)B+(\frac{(A^2+3B^2)A-1}{3B})A, \end{equation}

By having $\alpha, \beta$ and $\gamma$, you can back to the original $x,y,z,w$ numbers. An example of solutions can be found by setting $A=1$ and $B=1$ which leads to $\alpha=4,\gamma=1, \beta=5$ and after removing common factors $x=5$,$y=3$,$z=6$ and $w=-4$

\begin{equation} 3^3+4^3+5^3=6^3 \end{equation} which is similar to Pythagorean triplet $3^2+4^2=5^2$.

Arian
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The complete polynomial solution to

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

has been given by many people, so we need not duplicate that here. Your focus seem to be on the special case,

$$x_1^3+x_2^3 + x_3^3 = 1$$

Since this is a Diophantine problem, one does not really need elliptic curves since that is for rational solutions. But for integer solutions, what you need instead are Pell equations. Given,

$$(1-ac+bc)^3 + (a+c^2-ac^3)^3 + (ac^3-b-c^2)^3 = 1\tag1$$

its solution is,

$$(a,b,c,r) = (12qrt,\; 3(q-r)(3q+r)t,\; 3t^2,\; p-18qt^3)$$

and the Pell equation,

$$p^2-3(108t^6-1)q^2 = 1$$

Solution 1

We can use the trivial $(p,q) = (1,0)$. Substitute into the formula for $(a,b,c,r)$, and we get,

$$\color{blue}{(1 - 9 t^3)^3 + (9 t^4)^3 + (3t - 9 t^4)^3 = 1}$$

Let $t = -1$, and we have Ramanujan's famous taxicab number $10^3+9^3=12^3+1=1729.$

Solution 2

Then we use the Pell equation's non-trivial solution $(p,q) = (216t^6-1,\,\pm12t^3)$. Substitute into the formula for $(a,b,c,r)$ and using the positive case $q =+12t^3$ yields a deg-10 polynomial,

$$\small{\color{blue}{(1 - 9t^3 + 648t^6 + 3888t^9)^3 + (-135t^4 + 3888t^{10})^3 + (3t - 81t^4 - 1296t^7 - 3888t^{10})^3 = 1}}$$

Using the negative case $q = -12t^3$ yields a deg-16 polynomial.

Solution 3 and infinitely more

It is well-known that given the non-trivial solution $(p,q) = (216t^6-1,\,\pm12t^3)$ to a Pell equation, we can find an infinite more. Thus, substituting these into $(a,b,c,r)$, we can generate an infinite number of the blue formulas.

So not elliptic curves, but Pell equations since this is a Diophantine problem.

Tito Piezas III
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