# $1729$, Fermat's Last Theorem, and Ramanujan's sums of cubes formula

In the following page in one of Ramanujan's Lost Notebooks, Ramanujan found a formula for sums of cubes such as the famed $1729$.

Which can be found in the bottom right hand corner. But also in that page, is a formula:

If$$\sum_{n\geq0}a_nx^n=\frac {1+53x+9x^2}{1-82x-82x^2+x^3}\\\sum_{n\geq0}b_nx^n=\frac {2-26x-12x^2}{1-82x-82x^2+x^3}\\\sum_{n\geq0}c_nx^n=\frac {2+8x-10x^2}{1-82x-82x^2+x^3}$$ Then$$a_n^3+b_n^3=c_n^3+(-1)^n$$

My Question: How would you go about proving this?

I did notice one thing: It seemed like Ramanujan was using generating functions. Where given a sequence, you pretend that the numbers are coefficients of a polynomial, and you can then collapse that down into a single expression.

For example: The counting numbers $1,2,3,4,5,\ldots$ can be represented as$$1+2x+3x^2+4x^3+\ldots=\frac {1}{(1-x)^2}\tag{1}$$ But if you use generating functions, then you can't substitute that $x$ with anything, which makes them both powerful, but dangerous. So what would be the point of using generating functions?

• @DietrichBurde How are derivatives dangerous? – Frank Nov 6 '16 at 18:00
• I don't think "dangerous" has a well-defined meaning here (in particular, not in the original post). What question are you asking? Are you asking how to prove those identities of Ramanujan, or are you asking what the point of generating functions is? The latter question has already been answered on this site, I'm confident. As for the former question, you can use the given generating functions (and partial fractions, etc.) to find closed-form formulas for $a_n,b_n,c_n$, which probably leads to a direct verification. – Greg Martin Nov 6 '16 at 18:10
• @GregMartin Mainly, the point of this post is to prove the theorem. The other question (the point of generating functions) is just a secondary question that you don't have to answer. – Frank Nov 6 '16 at 18:21

For a short summary see also here. In fact, the existence of this result by Ramanujan depends very much on special circumstances of the solution $$(A^2 +7AB−9B^2)^3 +(2A^2 −4AB+12B^2)^3 = (2A^2 +10B^2)^3 +(A^2 −9AB−B^2)^3,$$ for $4$ cubes. Tito Piezas has found several other Ramanujan-like families of such identities with four cubes, see this MSE-question.