Pythagorean Triangle whose Area is Half Perimeter

Theorem

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:

$\AA = m n \paren {m^2 - n^2}$

The perimeter of $T$ is given by:

$\PP = m^2 - n^2 + 2 m n + m^2 + n^2 = 2 m^2 + 2 m n$

We need to find all $m$ and $n$ such that $\PP = 2 \AA$.

Thus:

$\ds 2 m^2 + 2 m n$

$=$

$\ds 2 m n \paren {m^2 - n^2}$

$\ds \leadsto \ \ $

$\ds m + n$

$=$

$\ds n \paren {m + n} \paren {m - n}$

$\ds \leadsto \ \ $

$\ds n \paren {m - n}$

$=$

$\ds 1$

As $m$ and $n$ are both (strictly) positive integers, it follows immediately that:

$n = 1$

$m - n = 1$

and so:

$m = 2, n = 1$

and the result follows.

$\blacksquare$