139. Find a right triangle in which the hypotenuse is expressed by the same number as the area.
139. Bhaskara uses the identity:

$$
\left[\left(m^{2}+n^{2}\right) x\right]^{2}=\left[m^{2}-n^{2}, x\right]^{2}+\left(2 m n x\right)^{2}
$$

and takes $\left(m^{2}+n^{2}\right) x$ as the hypotenuse, and $\left(m^{2}-n^{2}\right) x$ and $2 m n x$ as the legs. Then, by the condition $\left(m^{2}+n^{2}\right) x=m n\left(m^{2}-n^{2}\right) x^{2}$, it follows that,

$$
x=\frac{m^{2}+n^{2}}{m n\left(m^{2}-n^{2}\right)}
$$