8.5. Determine all ordered pairs of natural numbers $(x, y)$ that satisfy the equation

$$
\left|4 x^{2}-4 x y+y^{2}-3 x-2 y+56\right|+3 x+2 y=56
$$
Solution. The equality from the statement is written as follows: $\left|(2 x-y)^{2}-(3 x+2 y-56)\right|=-(3 x+2 y-56)$. Let $m=3 x+2 y-56, 2 x-y=n$. We obtain $-m=\left|n^{2}-m\right| \geq 0 \Rightarrow m \leq 0 \Rightarrow n^{2}-m \geq 0$.

It follows that $n^{2}-m=-m \Rightarrow n=0 \Rightarrow 2 x-y=0 \Rightarrow y=2 x$. Switching to the variable $x$, we get the equality $|56-7 x|=56-7 x \Leftrightarrow 56-7 x \geq 0 \Leftrightarrow x \leq 8$. Since $x \in \mathbf{N}, x \leq 8, y=2 x$, we get $S=\{(x, y)\}=\{(x, 2 x) \mid x \in\{0,1,2, \ldots, 8\}\}=\{(0,0),(1,2),(2,4), \ldots,(8,16)\}$, a total of 9 pairs of numbers.