Let's find the integer solutions for \(x\) and \(y\) in the following equation:

$$
2 x^{2}+8 y^{2}=17 x y-423
$$
I. solution. It is easy to see that if $\left(x_{0}, y_{0}\right)$ is a solution to (1), then $\left(-x_{0},-y_{0}\right)$ is also a solution, so it is enough to find those solutions where $y>0$ (since $y=0$ is not a solution). Consider $y$ as a parameter for the moment, so from the equation

$$
2 x^{2}-17 y x+\left(8 y^{2}+423\right)=0
$$

we get

$$
x=\frac{1}{4}(17 y \pm \sqrt{D})
$$

where $D$ denotes the discriminant:

$$
D=289 y^{2}-8\left(8 y^{2}+423\right)=225 y^{2}-3384
$$

this must be a perfect square of some integer $z$, and the expression in (2) must be divisible by 4. Thus

$$
225 y^{2}-z^{2}=(15 y-z)(15 y+z)=3384=2^{3} \cdot 3^{2} \cdot 47
$$

so $15 y-z=d_{1}$ and $15 y+z=d_{2}$ are related divisors of 3384:

$$
\begin{aligned}
& 15 y-z=d_{1} \\
& 15 y+z=d_{2}
\end{aligned}
$$

and from this, taking (2) into account as well:

$$
\begin{gathered}
y=\frac{d_{1}+d_{2}}{30}, \quad z=\frac{d_{2}-d_{1}}{2} \\
x_{1}=\frac{1}{4}(17 y+z)=\frac{d_{1}+16 d_{2}}{60}, \quad x_{2}=\frac{1}{4}(17 y-z)=\frac{16 d_{1}+d_{2}}{60}
\end{gathered}
$$

Because of $z$, $d_{1}$ and $d_{2}$ must have the same parity, i.e., both must be even. For $y$ to be an integer, both must also be divisible by 3, since if one is not divisible by 3, the other must be divisible by $3^2$, and thus $d_{1}+d_{2}$ would not be divisible by 3. Therefore, $d_{1}$ and $d_{2}$ are divisible by 6, and their other factors can only be chosen from the factors of $3384: 6^{2}=94=2 \cdot 47$. There are two possibilities (since $d_{1}, d_{2}$ have the same sign, and $y>0$ implies they are positive):

$$
\operatorname{or}\left\{\begin{array}{l}
d_{1}=6 \cdot 1, \\
d_{2}=6 \cdot 94,
\end{array} \quad \quad \operatorname{or} \left\{\begin{array}{l}
d_{1}=6 \cdot 2 \\
d_{2}=6 \cdot 47
\end{array}\right.\right.
$$

(exchanging them would only change the sign of $z$, which is irrelevant).

From the second, $y$ is not an integer; from the first,

$$
y=19, \quad z=279, \quad x_{1}=11, \quad x_{2} \text { is not an integer }
$$

therefore, (1) has only two solutions in integers:

$$
x=11, \quad y=19 \quad \text { and } \quad x=-11, \quad y=-19 \text {. }
$$

II. solution. We try to factorize the expression containing the unknowns. Experimenting with undetermined coefficients, from the desired

$$
2 x^{2}-17 x y+8 y^{2}=(A x+B y)(C x+D y)
$$

identity, by comparing coefficients, we get the equations

$$
A C=2, \quad B D=8, \quad B C+A D=-17
$$

According to these, the product of the numbers $B C, A D$ is 16, their sum is -17, so these products satisfy the equation

$$
z^{2}+17 z+16=0
$$

Thus, one of $B C, A D$ is -1, the other is -16. According to $A C=2$, if the coefficients are sought among the integers, one of $A$ and $C$ is 1, the other is 2. If $A=2$ and $C=1$, then $A D$ can only be -16, so $D=-8$ and $B=-1$.

After factorization, we get that (1) is equivalent to the equation

$$
(2 x-y)(8 y-x)=9 \cdot 47
$$

The difference of the two factors on the left side is $(9 y-3 x)$, which is divisible by 3, so only two factorizations of the right side are possible: $3 \cdot 141$ and $(-3) \cdot(-141)$. If generally

$$
2 x-y=U \quad \text { and } \quad 8 y-x=V
$$

then

$$
x=\frac{8 U+V}{15}, \quad y=\frac{U+2 V}{15}
$$

Since $141+2 \cdot 3$ is not divisible by 5, $U= \pm 141$ does not lead to an integer solution. However, if $U= \pm 3$, then $x=11, y=19$ or $x=-11, y=-19$, so these two pairs of numbers are the only solutions.