Finding prime solutions to $100q+80 = p^3 + q^2$

Question

Finding prime solutions to $100q+80 = p^3 + q^2$

Does them being prime imply some patterns on division modulo 3 or some other integer? How is this done?

Answer 1

solving your equation for $q$ we obtain $q=50\pm\sqrt{2580-p^3}$ we get $p\le 13$ therefore $p\in \{2,3,5,7,11,13\}$

Answer 2

Outline: One can do it crudely. Rewrite the equation as $-p^3=q^2-100q-80$, and note that the smallest value of $q^2-100q-80$ is $-2580$.