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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 '11 at 22:46

2 Answers 2

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.

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More generally than the Jacobi theta function, one can attach a theta function to any quadratic form. The case of cubic forms is much much more difficult. Very roughly speaking, quadratic forms are easier because they are connect to bilinear forms, thus to inner product spaces and ultimately Fourier analysis. There are theta functions attached to cubic forms, see the work of Patterson for example. But they are a lot more abstract and it's difficult to use them to make concrete deductions such as you're looking for.

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