Square of Hypotenuse of Pythagorean Triangle is Difference of two Cubes/Refutation

Theorem
Let $h$ be the hypotenuse of a Pythagorean triangle.

Then it is not necessarily the case that:
 * $h^2 = a^3 - b^3$

for some $a, b \in \Z_{>0}$.

Refutation
Consider the equation:


 * $(1): \quad h^2 = a^3 - b^3: a, b \in \Z_{>0}$

$h$ itself cannot be a cube, as this would be a counterexample to Fermat's Last Theorem.

Suppose we relax $\Z_{>0}$ to $\Z_{\ge 0}$.

It can be shown that $(1)$ does not hold even among the primitive Pythagorean triples.

We demonstrate that there is no solution for $h = 5$:

there is.

We must have $\paren {a - b} \divides h^2$.

For $x \ge h$:
 * $\paren {x + 1}^3 - x^3 = 3 x^2 + 3 x + 1 > h^2$

Hence $a \le 5$.

Combined with our divisibility condition, we must have:
 * $a - b = 1$ or $5$

But:
 * $2^3 < 25$
 * $3^3 - 2^3 = 19$
 * $4^3 - 3^3 > 25$
 * $5^3 - 0^3 > 25$

So there is no solution for $a^3 - b^3 = 25$.