Prove that the equation $ax^2+by^2=z^3$ in integers has infinitely many solutions

Question

Let $a,b \in \mathbb Z$ and $\gcd(a,b)=1$.Prove that the equation $$ax^2+by^2=z^3$$ in integers has infinitely many solutions satisfying $ \gcd(x,y)=1$.

I have no idea how to solve this problem.

Answer 1

When $-ab$ is a square, such as $x^2 - y^2,$ the set of represented numbers is defined only by a handful of congruences. For $x^2 - y^2,$ all odd numbers and all multiples of $4$ are represented, but nothing $2 \pmod 4.$

when $-ab$ is not a square, the form $a x^2 + b y^2 $ represents infinitely many primes, for example. However, no real need for primes, just take $u,v$ with gcd one, $$ 1 = \gcd(a,b) = \gcd(u,v) = \gcd(u,a) = \cdots $$ Then the principal form $x^2 + ab y^2$ represents the square of $$ z = a u^2 + b v^2 $$ and Gauss composition of the principal form with this form gives back $z^3.$

Take me a few minutes to get formulas.

$$ z^3 = a \left( a u^3 - 3 b u v^2 \right)^2 + b \left( 3a u^2 v- b v^3 \right)^2 $$ Since the terms $a u^3 - 3 b u v^2$ and $3a u^2 v- b v^3$ are homogeneous of degree three, we may divide through by any common factor that may have been introduced (such as $2$) and still get a cube as a result.

In case the gcd business is a worry: if some prime $p$ divides any one of $a,b,u,v$ and $p \neq 2,3,$ then $p$ does not divide either of $a u^3 - 3 b u v^2$ and $3a u^2 v- b v^3.$ And so on...