Problem 9.6. Find all pairs $(x, y)$ of natural numbers that satisfy the equation

$$
x^{2}-6 x y+8 y^{2}+5 y-5=0
$$
Solution. The equation is arranged as a quadratic equation in terms of the unknown $x$:

$$
x^{2}-6 y \cdot x+\left(8 y^{2}+5 y-5\right)=0
$$

Its discriminant, $\Delta=4 y^{2}-20 y+20$ must be a perfect square. Let $4 y^{2}-20 y+20=k^{2}$; for $k$ natural values are sufficient. It follows that $(2 y-5)^{2}-k^{2}=5$, or $(2 y-5-k)(2 y-5+k)=5$. For $k$ natural, the inequality $2 y-5-k \leqslant 2 y-5+k$ holds. Since $5=1 \cdot 5=(-1) \cdot(-5)$, there are two possibilities for the factors on the left side:

$$
\text { 1. }\left\{\begin{array} { l } 
{ 2 y - 5 - k = 1 , } \\
{ 2 y - 5 + k = 5 ; }
\end{array} \quad \text { 2. } \left\{\begin{array}{l}
2 y-5-k=-5 \\
2 y-5+k=-1
\end{array}\right.\right.
$$

The first system has the solution $y=4, k=2$. Substituting $y=4$ into the equation from the statement, we get $x^{2}-24 x+143=0$, which leads to the solutions $(11,4)$ and $(13,4)$.

From the second system, we find $y=1, k=2$. For $y=1$, the equation from the statement takes the form $x^{2}-6 x+8=0$, from which we obtain the solutions $(2,1)$ and $(4,1)$.

Thus, the given equation has the set of solutions $S=\{(2,1),(4,1),(11,4),(13,4)\}$.

Remark. There are other methods of solving. For example, the equation from the statement can be written in the form $(x-3 y)^{2}=y^{2}-5 y+5$ and it is shown that $(y-3)^{2}<4$. For the remaining values, $y=0,1,2,3,4$, the solutions of the equation are found.