Three. (25 points) Given $m, n, p (m<n)$ are positive integers. Let $A(-m, 0), B(n, 0), C(0, p)$, and $O$ be the origin. If $\angle ACB=90^{\circ}$, and
$$
OA^2 + OB^2 + OC^2 = 3(OA + OB + OC),
$$
(1) Prove: $m+n=p+3$;
(2) Find the analytical expression of the quadratic function whose graph passes through points $A, B, C$.


(1) From $\angle A C B=90^{\circ}, O C \perp A B$, we get
$$
O A \cdot O B=O C^{2} \text {, }
$$

which means $m n=p^{2}$.
$$
\begin{array}{l}
\text { From } O A^{2}+O B^{2}+O C^{2} \\
=3(O A+O B+O C),
\end{array}
$$

we get
$$
\begin{array}{l}
m^{2}+n^{2}+p^{2}=3(m+n+p) . \\
\text { Also, } m^{2}+n^{2}+p^{2} \\
=(m+n+p)^{2}-2(m n+n p+m p) \\
=(m+n+p)^{2}-2\left(p^{2}+n p+m p\right) \\
=(m+n+p)^{2}-2 p(m+n+p) \\
=(m+n+p)(m+n-p),
\end{array}
$$

Thus, $m+n-p=3 \Rightarrow m+n=p+3$.
(2) From $m n=p^{2}, m+n=p+3$, we know that $m, n$ are

the two distinct positive integer roots of the quadratic equation
$$
x^{2}-(p+3) x+p^{2}=0
$$

Therefore, $\Delta=[-(p+3)]^{2}-4 p^{2}>0$.
Solving this, we get $-1<p<3$.
Since $p$ is a positive integer, $p=1$ or 2.
When $p=1$, the equation (1) becomes $x^{2}-4 x+1=0$, which has no integer solutions.

When $p=2$, the equation (1) becomes $x^{2}-5 x+4=0$, with roots $m=1, n=4$.

Let the quadratic function passing through points $A, B, C$ be
$$
y=k(x+1)(x-4) \text {. }
$$

Substituting the coordinates of point $C(0,2)$, we get
$$
\begin{array}{l}
2=k \times 1 \times(-4) \\
\Rightarrow k=-\frac{1}{2} .
\end{array}
$$

Therefore, the quadratic function passing through points $A, B, C$ is
$$
y=-\frac{1}{2}(x+1)(x-4)=-\frac{1}{2} x^{2}+\frac{3}{2} x+2
$$