10. Proof: Let $x, y$ be the lengths of the legs, and $z$ be the length of the hypotenuse, and $z-x=1$. By the Pythagorean theorem, we have
$$x^{2}+y^{2}=z^{2}$$

Since $z-x=1$, it follows that $(x, z)=1$. Therefore, by equation (6), it must be that $(x, y)=1$. According to the discussion in this chapter, the solutions $x$ and $y$ that satisfy (6) must include one odd and one even number. Since $z$ is odd and $z-x=1$, $x$ must be even, i.e., $2 \mid x$. Therefore, the solutions of (6) satisfy all the conditions of Theorem 2. By Theorem 2, the lengths of the sides can be expressed as
$$x=2 a b, \quad y=a^{2}-b^{2}, \quad z=a^{2}+b^{2}$$

where $a$ and $b$ are positive integers, $a>b$, $(a, b)=1$, and $2 \nmid(a+b)$. Since $z-x=1$, from (7) we get $a^{2}+b^{2}-2 a b=1$. Thus, $(a-b)^{2}=1$. Since $a>b$, we have $a-b=1$, i.e., $a=1+b$. Substituting this into (7), we get the formulas for the side lengths as
$$\begin{array}{l} x=2 a b=2(1+b) b=2 b^{2}+2 b \\ y=a^{2}-b^{2}=(1+b)^{2}-b^{2}=2 b+1 \\ z=a^{2}+b^{2}=(1+b)^{2}+b^{2}=2 b^{2}+2 b+1 \end{array}$$

where $b$ is any positive integer.