Pythagorean Triangles whose Area equal their Perimeter

Theorem
There exist exactly $2$ Pythagorean triples which define a Pythagorean triangle whose area equals its perimeter:


 * $(1): \quad \left({6, 8, 10}\right)$, leading to an area and perimeter of $24$


 * $(2): \quad \left({5, 12, 13}\right)$, leading to an area and perimeter of $30$.

Proof
From Area of Right Triangle, the area $\mathcal A$ is:
 * $\mathcal A = \dfrac {a b} 2$

where $a$ and $b$ are the legs.

$(1): \quad$ The area of the $\left({6, 8, 10}\right)$ triangle is $\dfrac {6 \times 8} 2 = 24$.

Its perimeter equals $6 + 8 + 10 = 24$.

$(2): \quad$ The area of the $\left({5, 12, 13}\right)$ triangle is $\dfrac {5 \times 12} 2 = 30$.

Its perimeter equals $5 + 12 + 13 = 30$.

It remains to prove that these are the only ones.

Let $a, b, c$ be the lengths of the sides of a Pythagorean triangle $T$.

Thus $a, b, c$ form a Pythagorean triple.

By definition of Pythagorean triple, $a, b, c$ are in the form:
 * $2 m n, m^2 - n^2, m^2 + n^2$

We have that $m^2 + n^2$ is always the hypotenuse.

Thus the area of $T$ is given by:
 * $\mathcal A = m n \left({m^2 - n^2}\right)$

The perimeter of $T$ is given by:
 * $\mathcal P = m^2 - n^2 + 2 m n + m^2 + n^2 = 2 m^2 + 2 m n$

We need to find all $m$ and $n$ such that $\mathcal P = \mathcal A$.

Thus:

As $m$ and $n$ are both (strictly) positive integers, it follows immediately that either:
 * $n = 1$
 * $m - n = 2$

and so:
 * $m = 3, n = 1$

leading to the triangle:
 * $a = 6, b = 8, c = 10$

or:


 * $n = 2$
 * $m - n = 1$

and so:
 * $m = 3, n = 2$

leading to the triangle:
 * $a = 12, b = 5, c = 13$

and the result follows.