Solvability of the Diophantine equation $x^{2} - y^{2} = 4z^{n}$? It is known that for every integer $z$ there are integers $x, y$ such that $x^{2} - y^{2} = z^{3}.$ In fact, given an integer $z$, taking $x := z(z+1)/2$ and $y := z(z-1)/2$ suffices.
But how is the integer solvability of the equation $x^{2} - y^{2} = 4z^{n}$ when $n \geq 5$? By intuition it seems that this Diophantine equation is unsolvable.

Every integer number $n$ that is not of the form $4k+2$ can be expressed as the difference of two squares. For instance:
$$ x= z^n+1, \quad y=z^n-1 $$
is always a solution of:
$$ x^2-y^2 = 4z^n. $$
