Solve the following equation if $x$ and $y$ are non-negative integers:

$$
x^{2}+(x+y)^{2}=(x+9)^{2} \text {. }
$$
After squaring, we arrive at the equation

$$
x^{2}+2 x y+y^{2}-18 x-81=0
$$

Rearranging this in terms of powers of $x$,

$$
x^{2}+2(y-9) x+y^{2}-81=0
$$

is obtained, and after completing the square,

$$
(x+y-9)^{2}=9(18-2 y)
$$

Since the left side of (2) is a perfect square, equality in (2) can only hold if $18 - 2 y$ is a perfect square, as its coefficient, 9, is also a perfect square. Given that $y \geqq 0$, the possible values of $18-2 y$ are the even perfect squares not greater than 18, which are 0, 4, and 16. The corresponding values of $y$ are then 9, 7, and 1.

The value of $x$ can then be obtained from the quadratic equation derived from (2). If $y=9$, then $x=0$; if $y=7$, then $x=8$ or $x=-4$; and if $y=1$, then $x=20$ or $x=-4$.

Considering that $x$ and $y$ are non-negative integers, we find that equation (1) has three solutions. These are

$$
\begin{array}{lll}
x_{1}=0, & x_{2}=8, & x_{3}=20 \\
y_{1}=9, & y_{2}=7, & y_{3}=1
\end{array}
$$

Since our transformations were equivalent, the solutions found indeed satisfy (1).