There’s Only One Inradius For This? (Take Two)

The sides of a primitive Pythagorean triple can be written as $x^2-y^2$ , $2xy$ and $x^2+y^2$ , where $x$ and $y$ are positive integers satisfying

• $x>y$
• $\gcd(x,y)=1$
• $x-y$ is odd

The inradius of a right-angled triangle with shorter sides $a,b$ and hypotenuse $c$ is given by $r=\frac{a+b-c}{2}$ ; in this case we have $r=y(x-y)$

Assume $r$ can be written in the form $r=hp^k$ , where $p$ is an odd prime that doesn't divide $h$ .

Then both $y=h$ , $x-y=p^k$ and $y=hp^k$ , $x-y=1$ provide valid solutions to the above system of equations.

Since we want an $r$ that gives exactly one solution, this $r$ must have no odd prime factors; in other words, it has to be a power of $2$ . The only power of $2$ in the interval is $\boxed{1024}$ , and the only triangle with this inradius corresponds to $y=1024$ , $x-y=1$ .