Before we formally begin the mathematics, let us share an anecdote about the famous Indian mathematical genius, Srinivasa Ramanujan. When Ramanujan was seriously ill, G.H. Hardy went to visit him. Hardy said, "I came in a taxi with the license plate number 1729, which seemed rather dull; I hope it isn’t an unlucky omen." Ramanujan replied, "No, it is a very interesting number. Among the numbers that can be expressed as the sum of two cubes in two different ways, 1729 is the smallest." (That is, 1729 = 1^3+12^3 = 9^3+10^3; such numbers were later called Taxicab numbers.) Littlewood responded to this anecdote by saying, "Every integer is a personal friend of Ramanujan." (Source: Wikipedia)
From this anecdote, we find that there indeed exist certain integers that can be expressed as two different sums of cubes. In other words, the Diophantine equation: x^3+y^3=z^3+w^3 has integer solutions (x,y,z,w) where the values are pairwise distinct (positive or negative). With the help of Eisenstein integers, we can derive an expression for the general solution. The general solution obtained here is more complex than those found in existing textbooks, but it allows us to obtain all integer solutions from integer parameters. (Simplified formulas exist that can also yield all integer solutions, but to obtain certain integer solutions, fractional parameters are required, which raises questions about how to generate all integer solutions systematically.)
First, we factorize both sides within the Eisenstein integers: (x+y)(x+y\omega)(x+y\omega^2)=(z+w)(z+w\omega)(z+w\omega^2) Let x+y\omega be denoted as \xi, and z+w\omega as \eta. We obtain: \xi\bar{\xi}\left(\omega\xi+\omega^2\bar{\xi}\right)=\eta\bar{\eta}\left(\omega\eta+\omega^2\bar{\eta}\right) A brief analysis shows that \xi must be a composite number in the ring of Eisenstein integers. Let us assume: \xi=\alpha\beta Then the left side of the equation becomes: \alpha\beta\bar{\alpha}\bar{\beta}\left(\omega\alpha\beta+\omega^2\bar{\alpha}\bar{\beta}\right) Note that the term in the parentheses must be a real number. Therefore, in general, we can set (\lambda is a real number): \omega\alpha\beta+\omega^2\bar{\alpha}\bar{\beta}=\lambda\gamma\bar{\gamma} Then the left side is: \alpha\beta\bar{\alpha}\bar{\beta}\left(\lambda\gamma\bar{\gamma}\right) According to the characteristics of the right side, it should be similar to the left side. Thus, the factors on the right side must be a rearrangement of the factors on the left. Without loss of generality, one rearrangement is: \alpha\gamma\bar{\alpha}\bar{\gamma}\left(\lambda\beta\bar{\beta}\right) That is to say: \left\{\begin{aligned}&\eta=\alpha\gamma\\ &\bar{\eta}=\bar{\alpha}\bar{\gamma}\\ &\omega\eta+\omega^2\bar{\eta}=\lambda\beta\bar{\beta}\end{aligned}\right. Combining this with the known: \left\{\begin{aligned}&\xi=\alpha\beta\\ &\bar{\xi}=\bar{\alpha}\bar{\beta}\\ &\omega\xi+\omega^2\bar{\xi}=\lambda\gamma\bar{\gamma}\end{aligned}\right. We observe that: \left\{\begin{aligned}&\omega\alpha\gamma+\omega^2\bar{\alpha}\bar{\gamma}=\lambda\beta\bar{\beta}\\ &\omega\alpha\beta+\omega^2\bar{\alpha}\bar{\beta}=\lambda\gamma\bar{\gamma}\end{aligned}\right. Now comes the crucial step: we reverse our perspective and treat these as a system of equations for \alpha and \bar{\alpha}. This is simply a system of two linear equations with two variables! Solving it yields: \alpha=\lambda\omega^2\frac{\beta\bar{\beta}^2-\gamma\bar{\gamma}^2}{\gamma\bar{\beta}-\beta\bar{\gamma}}
In this way, we have completed the construction of the solution: given parameters \lambda, \beta, \gamma, we can calculate \alpha, and then according to: \xi=\alpha\beta,\quad\eta=\alpha\gamma we can find \xi and \eta, and subsequently determine x, y, z, w. Attentive readers will notice: \lambda is a real number, and \beta, \gamma are Eisenstein integers. From a real number perspective, \beta and \gamma each carry two free parameters, so we have obtained a solution with five free parameters! There are only four unknowns, yet there are five parameters; in principle, there is redundancy. However, in order to obtain all integer solutions from integer parameters, it is necessary to take it this way, which is quite wonderful.
Let \beta=a+b\omega and \gamma=c+d\omega. With the help of Mathematica, we can derive (where \lambda is a trivial parameter and is omitted): \left\{\begin{aligned}x&=\frac{a^3 (c+d)-3 a^2 b c+3 a b^2 c+b^3 (d-2 c)-(c^2-c d+d^2)^2}{3 b c-3 a d}\\[2ex] y&=\frac{a^3 (c-2 d)+3 a^2 b d-3 a b^2 d+b^3 (c+d)-(c^2-c d+d^2)^2}{3 a d-3 b c}\\[2ex] z&=\frac{a^4-2 a^3 b+3 a^2 b^2-a (2 b^3+c^3+d^3)+b (b^3+2 c^3-3 c^2 d+3 c d^2-d^3)}{3 a d-3 b c}\\[2ex] w&=\frac{-a^4+2 a^3 b-3 a^2 b^2+a (2 b^3+c^3-3 c^2 d+3 c d^2-2 d^3)+b (-b^3+c^3+d^3)}{3 a d-3 b c} \end{aligned}\right. A complex parametric solution...
When a=-1, b=-2, c=3, d=3, it gives x=9, y=-12, z=1, w=-10, which is 1^3+12^3=9^3+10^3, the smallest integer solution.
Perhaps interested readers would like to try finding solutions for x^4+y^4=z^4+w^4. Indeed, this equation also has pairwise distinct integer solutions; however, it is far more difficult than the cubic case. To date, we do not even know any form of a general solution for it.
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