One, (20 points) If the two sides of a right triangle, $x, y$, are both prime numbers, and make the algebraic expressions $\frac{2 x-1}{y}$ and $\frac{2 y+3}{x}$ both positive integers, find the inradius $r$ of this right triangle.
(1) If $x>y$, then
$$
1 \leqslant \frac{2 y+3}{x}<\frac{2 x+3}{x}<4 \text {. }
$$

It is easy to see that $\frac{2 y+3}{x}=1$ or 2.
$$
\begin{array}{l}
\text { (i) From } \frac{2 y+3}{x}=1 \\
\Rightarrow x=2 y+3 \\
\Rightarrow \frac{2 x-1}{y}=\frac{2(2 y+3)-1}{y}=4+\frac{5}{y} \\
\Rightarrow y=5, x=13 ;
\end{array}
$$
(ii) From $\frac{2 y+3}{x}=2 \Rightarrow 2 x=2 y+3$, the equation is obviously not valid, leading to a contradiction.
(2) If $x=y$, then $\frac{2 x-1}{y}$ is not a positive integer, which contradicts the given condition.
(3) If $x<y$, then
$$
\begin{array}{l}
1 \leqslant \frac{2 x-1}{y}<2 \Rightarrow \frac{2 x-1}{y}=1 \Rightarrow y=2 x-1 \\
\Rightarrow \frac{2 y+3}{x}=\frac{2(2 x-1)+3}{x}=4+\frac{1}{x}
\end{array}
$$

It cannot be a positive integer, which contradicts the given condition.
In summary, only $y=5, x=13$ meets the conditions of the problem. The third side of the right triangle is 12 or $\sqrt{194}$.
$$
\text { Hence } r=\frac{5+12-13}{2}=2
$$

or $r=\frac{5+13-\sqrt{194}}{2}=\frac{18-\sqrt{194}}{2}$.