If $k$ is an odd number then $3k^2 +16$ is not a perfect cube

Question

Show that if $k$ is odd then $3k^2+16$ is not a perfect cube.

Could anybody please prove it? I am particularly interested in a proof that mostly relies on divisibility.

Answer 1

We are looking for integral solutions to $3k^2+16=m^3$, with $k$ odd. Substituting $k$ and $m$ by $$k:=\frac{8}{9}y+\frac{4}{9}\qquad\text{ and }\qquad m:=\frac{4}{3}x,$$ and multiplying through by $\frac{27}{64},$ we find that any integral solution yields a rational solution to $$y^2+y=x^3-7,$$ which just so happens to be an equation for an elliptic curve over $\Bbb{Q}$, in minimal Weierstrass form.

This elliptic curve over $\Bbb{Q}$ has two nontrivial points, which are $(3,4)$ and $(3,-4)$, corresponding to the integral solutions $(k,m)=(4,4)$ and $(k,m)=(-4,4)$. These are then the only integral solutions to the original equation, and in particular there are no solutions if $k$ is odd.