25 Find all Pythagorean triples $(x, y, z)$ such that $x<y<z$, and $x, y, z$ form an arithmetic sequence.
25. From the conditions, we know $x^{2}+y^{2}=z^{2}$ and $x+z=2 y$, then
$$y^{2}=(z-x)(z+x)=2 y(z-x)$$

Thus,
$$y=2(z-x)$$

Therefore, $y$ is even. Let $y=2 m$, then

we get
$$\begin{array}{c}
x+z=4 m, z-x=m \\
x=\frac{3 m}{2}, z=\frac{5 m}{2}
\end{array}$$

Hence, $m$ is even. Therefore,
$$(x, y, z)=(3 n, 4 n, 5 n)$$

where $n=\frac{m}{2}$ is a positive integer.