$x^2 + ax - b=0\Longrightarrow x_{1,2}=\dfrac{-a\pm\sqrt{a^2+4b}}{2}\in\mathbb{Z}\Longrightarrow$
$\Longrightarrow \sqrt{a^2+4b}\in\mathbb{N}\Longrightarrow a^2+4b=m^2, m\in\mathbb{N}$.
$x^2 - ax + b=0\Longrightarrow x_{3,4}=\dfrac{a\pm\sqrt{a^2-4b}}{2}\in\mathbb{Z}\Longrightarrow$
$\Longrightarrow \sqrt{a^2-4b}\in\mathbb{N}\cup\{0\}\Longrightarrow a^2-4b=n^2, n\in\mathbb{N}\cup\{0\}$.
$m>n; m$ and $n$ have the same parity, hence $\dfrac{m+n}{2}=u\in\mathbb{N},\dfrac{m-n}{2}=v\in\mathbb{N}$.
From $\begin{cases}a^2+4b=m^2\\a^2-4b=n^2\end{cases}$ results:
$\begin{cases}a^2=\dfrac{m^2+n^2}{2};\\b=\dfrac{m^2-n^2}{8}.\end{cases}$
We can write:
$a^2=\left(\dfrac{m+n}{2}\right)^2+\left(\dfrac{m-n}{2}\right)^2=u^2+v^2$.
$b=\dfrac{1}{2}\cdot\dfrac{m+n}{2}\cdot\dfrac{m-n}{2}=\dfrac{uv}{2}$.
The right triangle with the legs $u$ and $v$ has the hypotenuse $\sqrt{u^2+v^2}=a$ and the area $A=\dfrac{uv}{2}=b$.