The general Taxicab equation is:

To eliminate duplicates and trivial solutions, the additional constraints $a \neq c, a \geq b, c \geq d$ are usually added. The history of this problem is nicely summarized on Wikipedia. For $k = 3$ the sequence is known as A001235, for $k = 4$ as A018786 and $k = 5$ is conjectured to be impossible. For the latter Randy L. Ekl showed computaionally that no solution $a^5 + b^5 = c^5 + d^5 \leq 4.01 \cdot 10^{30}$ exists. Using this solver I was able to extend this result and demonstrate that no solution exists $\leq 10^{34}$.

To solve this equation algorithmically, an upper bound $N$ with $a, b \leq N$ is chosen. One then constructs a table with value $a^k + b^k$ for all combinations $(a, b)$ with $a \geq b$ and then searches for duplicates. Each duplicate has at least two representations as the sum $a^k + b^k$ and is therefore a taxicab number. The memory complexity of this approach is $O(N^2)$ for the table, which means a memory requirement of hundreds of gigabytes already for $N = 10^6$.

In Enumerating solutions to p(a) + q(b) = r(c) + s(d) Daniel J. Bernstein describes a solution to this problem. He proposes to replace the table with a minimal heap that is iteratively updated and therefore contains only $O(N)$ values at each time point. This also allows us to enumerate all combinations $(a, b)$ such that the result $a^k + b^k$ is in ascending order. In this way, we can easily find duplicate values as they succeed each other directly. Further details can be found in section 3 of the paper.

(c) Mia Müßig