Any solution in positive integers of the equation $y^3 -x^3 = z^2$, other than $\left(x, y, z\right) = \left(7, 8, 13\right)$?

Question

Solve in positive integers: $$ y^3 - x^3 = z^2,$$ where $\gcd (x, y, z)=1$.

Answer 1

Yes, there are many such examples. A quick search on Mathematica gives the following to be solutions (not exhaustive, of course): $$(x,y,z)=(23,71,588),(47,74,549),(118,193,2355),(157,349,6216).$$ There are more solutions like $$(x,y,z)=(28,56,392),(31,155,1922),(119,140,1029),(55,176,2299),$$ which do not satisfy the $\gcd(x,y,z)=1$ condition. As far as I know, there is probably no easy way to classify all solutions, since the given equation does not fall into any well-known type, and in general Diophantine equations of order $\geq 3$ for more than $2$ variables are not very well-understood. Even understanding cubic equations for $2$ variables is already quite a challenge, as a quick search for "Elliptic Curves" will surely reveal.

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Edit: There is actually a way to paramatrise all the solutions to the given equation, which I didn't know when I wrote the answer. Refer to Mike Bennett's answer for details!

Answer 2

Actually, it is relatively straightforward to parametrize all solutions of the equation $x^3-y^3=z^2$ in coprime integers. This is done, for example, in Section 14.3 of Henri Cohen's Springer GTM 240; there are three parametrized families of solutions. By way of example, one family has $$ x=-3s^4+6s^2t^2+t^4, \; y=-3s^4-6s^2t^2+y^4 $$ and $z=6st(3s^4+t^4)$, with $s$ and $t$ integers of opposite parity, and $3 \nmid t$.