Solve $z^3=kx+ny$ , ($k\neq{n},k,n\in \mathbb{N}$)

Question

Solve $z^3=kx+ny$ , ($k\neq{n},k,n\in \mathbb{N}$) for positive integer unknowns $x,y,z$

I have really no idea for this!!

Answer 1

Suppose $g = \gcd(k,n)$. There are integers $a,b$ such that $ak + bn = g$. Let's suppose $a > 0$ and $b < 0$, and let $s = k/g$, $t = n/g$. If $z^3 = u g$ is a cube divisible by $g$, then for any integer $j$ you can write

$$ z^3 = (u a - j t) k + (u b + j s) n $$

You want $x = u a - j t > 0$ and $y = u b + j s > 0$, i.e. $$ -\dfrac{ub}{s} < j < \dfrac{ua}{t}$$ Since $a/t + b/s = 1/(st) > 0$, this is possible as long as $u$ is sufficiently large.