# Inradius of Pythagorean Triangle is Integer/Proof 1

## Theorem

Let $\triangle ABC$ be a Pythagorean triangle.

Let $r$ be the inradius of $\triangle ABC$.

Then $r$ is an integer.

## Proof

Let $\triangle ABC$ be such that $\angle C$ is a right angle.

Let:

$A$ be opposite the side $a$
$B$ be opposite the side $b$
$C$ be opposite the side $c$ From Solutions of Pythagorean Equation, we have that:

$c = m^2 + n^2$

for some $m, n \in \Z_{>0}$, and that the other sides are $m^2 - n^2$ and $2 m n$

Without loss of generality, let the sides $a$ and $b$ be such that:

 $\displaystyle a$ $=$ $\displaystyle m^2 - n^2$ $\displaystyle b$ $=$ $\displaystyle 2 m n$

Hence:

 $\displaystyle c$ $=$ $\displaystyle \paren {a - r} + \paren {b - r}$ $\displaystyle$ $=$ $\displaystyle a + b - 2 r$ $\displaystyle \leadsto \ \$ $\displaystyle r$ $=$ $\displaystyle \dfrac {a + b - c} 2$ $\displaystyle \leadsto \ \$ $\displaystyle r$ $=$ $\displaystyle \dfrac {m^2 - n^2 + 2 m n - \paren {m^2 + n^2} } 2$ $\displaystyle \leadsto \ \$ $\displaystyle r$ $=$ $\displaystyle \dfrac {2 m n - 2 n^2} 2$ $\displaystyle \leadsto \ \$ $\displaystyle r$ $=$ $\displaystyle n \paren {m - n}$

As $m$ and $n$ are both integers, it follows that $r$ is also an integer.

$\blacksquare$