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

Consider the equation:
 * $h^2 = a^3 - b^3, a, b \in \Z_{>0}$

If $h$ is a cube, a solution to the above would violate Fermat's Last Theorem.

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

Even among the primitive Pythagorean triples, we can show that there is no solution for $h = 5$:


 * Suppose 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 $a^3 - b^3 = 25$.

Refer to A038597.

It boils down to which numbers in that sequence are hypotenuses, which relates to sum of two squares.

(Probably numbers that, after removing their $4n+3$ factors, is not a square or twice a square)

After $5^2 + 12^2 = 13^2 = 8^3 - 7^3$, the next ones are $104, 181, 388, \dots$:


 * $96^2 + 40^2 = 104^2 = 32^3 - 28^3$
 * $19^2 + 180^2 = 181^2 = 105^3 - 104^3$
 * $260^2 + 288^2 = 388^2 = 114^3 - 110^3$

etc.--RandomUndergrad (talk) 05:43, 31 August 2020 (UTC)


 * Thank you for this, another black mark against Wells's name ... --prime mover (talk) 06:15, 31 August 2020 (UTC)