## Representations as a sum of cubes following Jacobi

Jacobi connected the generating function counting the number of representations as a square with elliptic trigonometry and use Fourier series to find the exact congruence condition and formula for counting representations as a sum of three squares [1].

To be precise it was the theta function $$1 + \sum_{n=1}^\infty 2 q^{n^2}$$

I was wondering if it was possible to use this approach on positive cubes $$\sum_{n=1}^\infty q^{n^3}$$ and integer cubes $$\sum_{n=-\infty \ldots \infty} q^{n^3}$$ since there is no useful algebraic object (like the Gaussian integers, quaternions and such) for cubes.

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No, it's not possible. But I read a spoof article (perhaps in the Notices of the AMS, perhaps on April 1) several years ago, about a great unrecognized mathematician who had made such discoveries. – paul Monsky Mar 7 2011 at 22:46

I'm sure the sums you write down are used, in conjunction with the circle method, to find asymptotic expressions for the number of representations as sums of cubes, but if it were possible to do for cubes what Jacobi et al did for squares we wouldn't be in the position of not knowing whether $33$ is a sum of three cubes.