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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?

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    $\begingroup$ given any $n = u^2 + p v^2,$ there is a formula for $x^2 + p y^2 = n^3.$ If the class number $h(-4p)$ is larger than $1$ there may be more; an example would be $n = 3 u^2 + 2uv + 4 v^2$ leading to $n^3 = x^2 + 11 y^2.$ $\endgroup$ – Will Jagy Feb 11 '19 at 1:33
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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.

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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.

$$x=p^6+q(b^2+8bs-5s^2)p^4+q^2(s^2-b^2)(b^2-8bs-5s^2)p^2+q^3(s^2-b^3)^3$$

$$y=2p(q^2(2s+b)(s^2-b^2)^2+2qb(b^2-3s^2)p^2-(2s-b)p^4)$$

$$z=p^4+2q(s^2+b^2)p^2+q^2(s^2-b^2)^2$$

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  • $\begingroup$ How did you figure out that solution? $\endgroup$ – StarCrunch Feb 11 '19 at 7:58
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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;

$302^2+(3)(1057)^2=(151)^3$

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