Square - square = cube

You might know that (1+2+3...n)^2=1^3+2^3+3^3...n^3.So (1+2+3...n)^2-(1+2+3...n-1)^2=n^3

Let us take $x^2 - y^2 = k^3,$ which is expressible as $(x + y)(x - y) = k^3.$ Now let us take the following system of equations:

$x + y = k^2; x - y = k \Rightarrow x = \frac{k^2 + k}{2} = \frac{k(k+1)}{2}; y = \frac{k^2 - k}{2} = \frac{k(k-1)}{2}.$

Hence, $x, y \in \mathbb{Z}$ for all $k \in \mathbb{Z}$ and conclusively shows all perfect cubes can be expressed as the difference of two perfect squares.

QED