10.4.1. (12 points) Find all pairs of integers $(x, y)$ that are solutions to the equation

$$
7 x y-13 x+15 y-37=0 \text {. }
$$

In your answer, indicate the sum of the found values of $x$.
Answer: 4.

Solution. Multiplying both sides of the equation by 7, we get

$$
7 \cdot 7 x y - 13 \cdot 7 x + 15 \cdot 7 y - 37 \cdot 7 = 0
$$

Rewriting this equation, we have

$$
(7 x + 15)(7 y - 13) = 64
$$

If $x$ and $y$ are integers, then the numbers $7 x + 15$ and $7 y - 13$ are also integers and leave a remainder of 1 when divided by 7. All possible factorizations of the number 64 into the product of two integers are as follows:

$$
64 = 1 \cdot 64 = 2 \cdot 32 = 4 \cdot 16 = 8 \cdot 8 = -1 \cdot (-64) = -2 \cdot (-32) = -4 \cdot (-16) = -8 \cdot (-8)
$$

Among these factors, the numbers that leave a remainder of 1 when divided by 7 are 1, 8, and 64. Therefore, the possible cases are

$$
\begin{gathered}
7 x + 15 = 1, 7 y - 13 = 64 \Rightarrow (x, y) = (-2, 11) \\
7 x + 15 = 8, 7 y - 13 = 8 \Rightarrow (x, y) = (-1, 3) \\
7 x + 15 = 64, 7 y - 13 = 1 \Rightarrow (x, y) = (7, 2)
\end{gathered}
$$

Thus, the sum of the possible values of $x$ is $-2 - 1 + 7 = 4$.