Find all $z$ such that $ au^2+2buv+cv^2=z^{2k+1}$

Question

Consider the quadratic equation $$ au^2+2buv+cv^2=z^{2k+1}$$ where $a,b,c,u,v,z$ are integers. Must $z$ have the same form? Please expand your answer. Thanks.

Answer 1

the form class group for discriminant $-188$ is $$ \langle 1,0,47 \rangle, \; \langle 3,2,16 \rangle, \; \langle 3,-2,16 \rangle, \; \langle 7,6,8 \rangle, \; \langle 7,6,-8 \rangle \; . $$

The prime $3$ is represented by the two evident forms, which are "opposites" in the traditional language; that is, they are inverses in the class group, under Gauss composition. However, to primitively represent $27,$ we get $7 u^2 + 6 uv + 8 v^2 = 27$ with $u=1,v=-2$

This is just Gauss composition.

Meanwhile, the only way to write $27 = 3 x^2 + 2 xy + 16 y^2$ is with $y=0, x=\pm 3$