One, (50 points) If $x, y$ are natural numbers, solve the equation: $x^{2}-2 x y+y^{2}+5 x+5 y=1500$.


One, Solution: $\because(x-y)^{2}+5(x+y)=1500, \therefore(x-y)^{2} \equiv 0(\bmod 5)$.
Let $x-y=5 a$ (assuming $x \geqslant y$), then we have $(x-y)^{2}+5(x+y)=25 a^{2}+5(x+y)=1500$,
which means $x+y+5 a^{2}=300 . \therefore x+y \equiv 0(\bmod 5)$.
Let $x+y=5 b$, then we have $5 a^{2}+5 b=300$, which is $a^{2}+b=60$.
$\because a, b \in \mathbf{N}, \therefore$ the ordered pairs of real numbers $(a, b)$ can be:
$$
(0,60),(1,59),(2,56),(3,51),(4,44),(5,35),(6,24),(7,11) \text {. }
$$

Therefore, the ordered pairs of real numbers $(x, y)=\left(\frac{5(a+b)}{2}, \frac{5(b-a)}{2}\right)$ can be:
$$
(150,150),(150,145),(145,135),(135,120),(120,100),(100,75),(75,45),(45,10) \text {. }
$$

Thus, the solutions $(x, y)$ of the original equation are 15 pairs, which are:
$$
\begin{array}{l}
(150,150),(150,145),(145,135),(135,120),(120,100),(100,75),(75,45),(45,10), \\
(145,150),(135,145),(120,135),(100,120),(75,100),(45,75),(10,45) .
\end{array}
$$