# On $x^2 + y^3 = z!$ with $y \ge 0$

The largest solution that I found to equation $$x^2 + y^3 = z!$$ with $$y \ge 0$$ as below.

$$6389296200^2 + 2173500^3 = 21!$$

It is obvious that if $$z!/t^{6} = x_{t}^2 + y_{t}^3$$ then $$z! = (x_{t}t^3)^2 + (y_{t}t^2)^3$$. So for a large value of $$z$$, there can be certain set of $$t$$ values to check existence of a smaller solution that provides $$x^2 + y^3 = z!$$ is soluble.

In that direction, I believe that there can be search to other solutions to this equation.

Question. Are there other solutions to $$x^2 + y^3 = z!$$ for $$z > 21$$? (It is checked up to $$33!$$).

I don't have any reference on that equation except https://oeis.org/A273553. So any reference is very welcome if this is well-known equation.

Thanks.

• Correct me if I'm wrong but technically you can't call this a Diophantine equation because of the factorial, surely? – JamalS Aug 11 at 12:16
• Heuristically, this should have finitely many solutions, since the numbers of numbers of the form $x^2 +y^3$ which are at most t, is $O(t^{5/6})$, and the set of factorials grows faster than exponential. I'd be surprised if there are any larger ones than the solution for 21!. On the other hand, I'm a bit surprised that that one exists at all. – JoshuaZ Aug 11 at 12:47
• From the perspective of elliptic curves, the restriction that $y \geq 0$ is somewhat unnatural (and removing it finds more solutions, for example with $n = 32$). If you remove this restriction, there are in fact $9$ representations of $21!$ in the form $x^{2} + y^{3}$ (including a second with $y > 0$: $5462360064^{2} + 2769984^{3}$). – Jeremy Rouse Aug 11 at 13:20