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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.

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Airdish
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    Well $x,y$ are variables, so they range over all relevant values, including the numbers in the post. Did you have an idea for a simpler substitution? –  Feb 22 '16 at 14:07
  • @PaulPlummer Not really, but the substitution just seems so out of the blue and complicated. So I feel that if it has to be that complicated, there must be some reason or constraints. Which is why I've asked the question. – Airdish Feb 22 '16 at 14:09
  • @PaulPlummer Also for the y substitution, the values are restricted since the substitution used has an even power, hence nothing below the vertex is considered. – Airdish Feb 22 '16 at 14:10

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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.

Giovanni Resta
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  • Nice. Do you know any simple but effective source where I can learn more about Diophantine Equations? I don't want anything completely exhaustive, just a few properties, methods of solving and so on... – Airdish Feb 22 '16 at 14:31
  • Well $(x,y,z)=(2n^3, 11n^3, 5n^2)$ would work too, as Calvin points out in the referred post. – Macavity Feb 22 '16 at 15:15