to answer to this kind of qestion we have to find a general formua and that is one of as formulas $(10+60n^{3})^{3}+(10-60n^{3})^{3}+(-60n^{2})^{3}+(-1)^{3}=1999$. Driss Dahbani.

How would one find this construction?

Well the idea is that variable terms have to cancel out in the expansion, and at the same time we must get an infinite sequence, so we try the easiest way: $ (a+bk)^3+(a-bk)^3= 2a^3 + 6ab^2k^2$, the next term should cancel that $ k^2$ out, so put $ k=n^3$ to get $ (a+bn^3)^3+(a-bn^3)^3+(-bn^2)^3=2a^3+(6a-b)b^2n^6$. If we put $ b=6a$, we're nearly there, since we have a constant. Now, $ 2a^3+c^3=1999$ should have a solution to be done. And indeed: $ a=10,c=-1$ is such a solution. Hope that explains the reasoning.

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"mbucko wrote: Nevertheless, I would like to point out one thing, i.e. generalization, e.g. $ A^n + B^n + C^n + D^n = 1999$. Haven't gone through this either Quote: Be careful, $ n$ has to be even otherwise all the terms are positive and limited by $ 1999$ so finite. In the trivial case $ n=1$ with two term is enough $ (1999+n)^1+(-n)^1=1999$ so probably $ a_1^{2k-1}+\cdots+a_{2k}^{2k-1}=m$ has infinite solutions.