<s>[INST] Diophantine equation. Three. Diophantine equation. $X^2+Y^2=qZ^3$
I wonder at what values ​​of the coefficient $q$ equation has a solution.
And of course I wonder how she looks like a formula describing their solutions.
For the special case when $X^2+Y^2=Z^3$ You can get a basic formula.
Has the solutions:
$X=2k^6+8tk^5+2(7t^2+8qt-9q^2)k^4+16(t^3+2qt^2-tq^2-2q^3)k^3+$
$+2(7t^4+12qt^3+6q^2t^2-28tq^3-9q^4)k^2+8(t^5+2qt^4-2q^3t^2-5tq^4)k+$
$+2(q^6-4tq^5-5q^4t^2-5q^2t^4+4qt^5+t^6)$
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$Y=2k^6+4(3q+t)k^5+2(9q^2+16qt+t^2)k^4+32qt(2q+t)k^3+$
$+2(-9q^4+20tq^3+30q^2t^2+12qt^3-t^4)k^2+4(-3q^5-tq^4+10q^3t^2+6q^2t^3+5qt^4-t^5)k-$
$-2(q^6+4tq^5-5q^4t^2-5q^2t^4-4qt^5+t^6)$
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$Z=2k^4+4(q+t)k^3+4(q+t)^2k^2+4(q^3+tq^2+qt^2+t^3)k+2(q^2+t^2)^2$
$q,t,k$ - What are some integers any sign.  After substituting the numbers and get a result it will be necessary to divide by the greatest common divisor. This is to obtain the primitive solutions.
 [/INST] 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.
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