Given prime $p$, find solutions to $x^2 + p y^2 = z^3$ For a prime $p$ consider non-zero integers $x,y,z$ that satisfy:
$$ x^2 + p y^2 = z^3$$
Does this fit in a known class of Diophantine equations that have been studied already?
I'm not sure how to go about solving these.  Looking at it mod $p$, it looks like it should be easy to just choose a $z$, cube it, check if the result is a quadratic residue and solve for $x$. But I'm not sure how to lift this to a solution in the integers.  Or maybe that is just a horrible starting approach.
How can I find solutions to this equation?
 A: If $z$ is of the form $x^2 + p y^2$, then so is any power of $z$, since 
$$ (a^2 + p b^2)(c^2 + p d^2) = (ac-pbd)^2 + p (ad + b c)^2 $$
If there is unique factorization in $\mathbb Z[\sqrt{-p}]$, those are all the solutions.
A: For the equation:  $$X^2+qY^2=Z^3$$
You can write this simple solution:
$$X=(p^2+qs^2)((p^4-q^2s^4)t^3-3(p^2+qs^2)^2kt^2+3(p^4-q^2s^4)tk^2-$$
$$-(p^4-6qp^2s^2+q^2s^4)k^3)$$
$$Y=2ps(p^2+qs^2)((p^2+qs^2)t^3-3(p^2+qs^2)tk^2+2(p^2-qs^2)k^3)$$
$$Z=(p^2+qs^2)((p^2+qs^2)t^2-2(p^2-qs^2)tk+(p^2+qs^2)k^2)$$
$q - $The ratio is given for the problem.
$p,s,t,k - $ integers asked us.
To describe the solutions of the equation.  $$x^2+qy^2=z^3$$
I think best would be to describe a solution using $3$ parameters.
$$x=p^6+q(b^2+8bs-5s^2)p^4+q^2(s^2-b^2)(b^2-8bs-5s^2)p^2+q^3(s^2-b^2)^3$$
$$y=2p(q^2(2s+b)(s^2-b^2)^2+2qb(b^2-3s^2)p^2-(2s-b)p^4)$$
$$z=p^4+2q(s^2+b^2)p^2+q^2(s^2-b^2)^2$$
A: Above equation shown below:
For $p=3$,
$x^2+3y^2=z^3$  ------$(A)$
Equation $(A)$ has solution:
$(x,y,z)=[(6m^2+8),(3m^3+4m),(3m^2+4)]$
For $m=7$ we get;
$302^2+(3)(1057)^2=(151)^3$
