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!
On
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.
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 $3^2+4^2=5^2$.