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

Source Work

1997: David Wells: Curious and Interesting Numbers (2nd ed.):

The Dictionary

$13$

This statement does not appear in the first edition of Curious and Interesting Numbers of $1986$.

Mistake

The square of the hypotenuse of a right-angled triangle is also the difference of $2$ cubes; thus, $13^2 = 8^3 - 7^3$.

Correction

This is not true for all right-angled triangles.

The most immediate counterexample is the $3-4-5$ triangle, whose hypotenuse is $5$.

The sequence of positive integers whose squares are the difference between $2$ cubes begins:

$13, 28, 49, 104, 147, 181, 189, 224, 351, 361, \dotsc$

This sequence is A038597 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).

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 $h = 5$:

Aiming for a contradiction, 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 $a^3 - b^3 = 25$.

$\blacksquare$

Sources

• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $13$