<s>[INST] Integer solutions of $x^3+y^3=z^2$ Is there any integer solution other than $(x,y,z)=(1,2,3)$ for $x^3+y^3=z^2$?
 [/INST] Here is the way to produce ALL integer solutions, not necessarily coprime, and not just some random infinite families.
For any integer $m$, let $D_2(m)$ be the set of integers $w$ such that $w^2$ is a divisor of $m$. For example, $D_2(18)=\{\pm 1,\pm 3\}$.
Then the set of all integer solutions to $x^3+y^3=z^2$ can be described as
$$
(x,y,z)=\left(\frac{u(u^3+v^3)}{w^2}, \, \frac{v(u^3+v^3)}{w^2}, \frac{(u^3+v^3)^2}{w^3} \, \right), \quad u,v \in {\mathbb Z}, \quad w \in D_2(u^3+v^3).
$$
Condition $w \in D_2(u^3+v^3)$ ensures that $x,y,z$ are integers, and a direct substitution proves that they satisfy the equation. Let us prove that, conversely, all integer solutions are covered by this family. Indeed, solution $(0,0,0)$ is covered. Let $(x,y,z)$ be any other solution, let $d$ be the greatest common square-free divisor of $x$ and $y$, and write $x=du$, $y=dv$ for some integers $u,v$. Then $d(u^3+v^3)=(z/d)^2$ is an integer, hence $z/d$ is an integer, say $z'$. Then $d \cdot (u^3+v^3)=(z')^2$ implies the existence of integers $a,b,w$ such that $d=ab^2$ and $u^3+v^3=aw^2$. Because $d$ is square-free, $b^2=1$, hence $d=a=\frac{u^3+v^3}{w^2}$. So, take $u,v$ arbitrary, $w \in D_2(u^3+v^3)$, and then express $x,y,z$. We obtain the stated formulas.
The method works for all equations of the form $F(x,y)=z^2$, where $F$ is a cubic form with integer coefficients.
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