How can any such substitution in a diophantine equation be possible?

Question

I saw this question and I do not understand the logic behind the $n$ substitution. Is it possible to substitute $x$ and $y$ with anything? If that is the case then you might is well have substituted $x$ and $y$ with something simpler, which still could factorized into a cube.

Answer 1

The question was "prove that $x^2+y^2=z^3$ has infinite non-trivial solutions".

One of the ways to prove such a thing is to prove that there is a family of solutions that derive from an identity.

In this case, the solution comes from finding that for every $n$ it holds $$ (n^3-3n)^2+(3n^2-1)^2=(n^2+1)^3\,. $$

Probably this is not the only family of solutions. But it is enough to prove that there are infinite solutions.

If your task was to find some polynomials in $n$ such that $p(n)^2+q(n)^2=r(n)^3$, the first thing you notice is that the degrees must match, and since the least common multiple of $2$ and $3$ is 6, then the simplest possible solution is when $r(n)$ is a quadratic polynomial and $p(n)$ and/or $q(n)$ is cubic.

Indeed, since the leading coefficients of $p(n)^2$ and $q(n)^2$ are both positive (because of the squaring) the degree of $p(n)^2+q(n)^2$ can't be smaller than twice the maximal degree between $p(n)$ and $q(n)$, because the leading monomials cannot cancel out.

This means that you can't have $r(n)$ linear and select two quadratic $p(n)$ and $r(n)$ such that $p(n)^2+q(n)^2$ magically is a cubic polynomial.

So, given the constraints on the degrees, I don't think there are simpler family of solutions.