Smallest Square Inscribed in Two Pythagorean Triangles

Theorem
The smallest square with integer sides that can be inscribed within two different Pythagorean triangles so that one side of the square lies on the hypotenuse has side length $780$.

The two Pythagorean triangles in question have side lengths $\tuple {1443, 1924, 2405}$ and $\tuple {1145, 2748, 2977}$.

Proof
For a Pythagorean triangles with side lengths $a, b, c$, the required inscribed square has side length given by:
 * $\dfrac {abc}{ab + c^2}$

For primitive pythagorean triples, $a, b, c$ are pairwise coprime, so the above fraction is in canonical form.

In other words, if the required side length is an integer, the triangle cannot be primitive, and this side length would be equal to some multiple of the product $abc$ of its primitive version.

Therefore in order to find the smallest such square, we would need to compare and find two sets of primitive pythagorean triples such that their product would have a sufficiently small lowest common multiple.

The $\tuple {3, 4, 5}$ triple has a product of $60$.

The $\tuple {5, 12, 13}$ triple has a product of $780$.

These two products have a lowest common multiple of $780$.

Since $\sqrt[3]{780} < 10$, there is no need to search any further for triangles with smallest side length greater than $10$.

The ones remaining are:
 * $7, 24, 25$
 * $8, 15, 17$
 * $9, 40, 41$

which all have products greater than $780$.

Therefore the solution must be generated with the two smallest triangles.

The inscribed square side length for the $\tuple {3, 4, 5}$ triangle is:
 * $\ds \frac {60}{3\times 4 + 5^2} = \frac {60}{37}$

so it must be enlarged $481$ times to have a side length of $780$.

The inscribed square side length for the $\tuple {5, 12, 13}$ triangle is:
 * $\ds \frac {780}{5\times 12 + 13^2} = \frac {780}{229}$

so it must be enlarged $229$ times to have a side length of $780$.

After enlargement, we get the triples $\tuple {1443, 1924, 2405}$ and $\tuple {1145, 2748, 2977}$, which is the result we have.