<s>[INST] write 33 as sum of three cubes to solve 3 as sum of three cubes I am interested in writing numbers as the sum of three positive or negative cubed integers. This article describes how a supercomputer was used to find that
$$33 = 8{,}866{,}128{,}975{,}287{,}528^3 + (−8{,}778{,}405{,}442{,}862{,}239)^3 + (−2{,}736{,}111{,}468{,}807{,}040)^3$$
Apparently they then used that to find a third way of writing 3 as the sum of three cubes - after $1^3+1^3+1^3$ and $4^3+4^3+(-5)^3$:
$$3 = 569{,}936{,}821{,}221{,}962{,}380{,}720^3 + (−569{,}936{,}821{,}113{,}563{,}493{,}509)^3 + (−472{,}715{,}493{,}453{,}327{,}032)^3$$
So our equation $33=x^3+y^3+z^3$ can be rewritten as $33-z^3=x^3+y^3$. Now, $x^3+y^3=(x+y)(x^2-xy+y^2)$. So for each $z$ values, we compute $33-z^3$ and then try to factor it and see if it can be reaarage into the form $(x+y)(x^2-xy+y^2)$. How else can we make improvement to reduce the size of search space to find a way to express 33 as sum of three cubes.  So how does finding 33 as sum of three cubes help find 3 as sum of three cubes?
 [/INST] The resolution for $33$ didn't help directly for the resolution for $3$. The initial author rather designed an efficient method to tackle the general problem.
In the paper (https://people.maths.bris.ac.uk/~maarb/papers/cubesv1.pdf), he describes an alternative method based on factorization: if
$$k-z^3=(x+y)(x^2-xy+y^2)$$ then you can try among the $x+y$ that divide $k-z^3$ and solve
$$\begin{cases}x+y=d,\\x^2-xy+y^2=\dfrac{k-z^3}d\end{cases}.$$
The method he used is different.
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