1. Find all integer solutions of the equation

$$
\frac{x^{2}}{2}+\frac{5}{y}=7
$$
II/1. If $(x, y)$ is a pair of numbers that solves the equation, then $(-x, y)$ also solves the equation. Therefore, it is sufficient to observe only those numbers $x$ that are non-negative. Multiplying the equation by $2 y$, we get $x^{2} y + 10 = 14 y$, from which we can express

$$
y = \frac{10}{14 - x^{2}}
$$

From this, it follows that $14 - x^{2}$ must be a divisor of the number 10, so $-10 \leq 14 - x^{2} \leq 10$ or $-24 \leq -x^{2} \leq -4$, hence $24 \geq x^{2} \geq 4$. Since we assumed that $x \geq 0$, we get $5 > x \geq 2$. We can calculate that for $x = 2$ we get $y = 1$, for $x = 3$ we get $y = 2$, and for $x = 4$ we get $y = -5$. Therefore, all integer solutions to the equation are the pairs $(2,1), (3,2), (4,-5)$, $(-2,1), (-3,2)$, and $(-4,-5)$.

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-09.jpg?height=69&width=1633&top_left_y=2284&top_left_x=220)

Observation that the number $x^{2} - 14$ or the number $y$ is a divisor of 10 ................................................... 1 point

Estimate $4 \leq x^{2} \leq 24$ or list the possibilities for divisors ................................................... 2 points

(For considering only positive divisors of the number 10, award 1 point)

Writing down the solutions ........................................................................ (For 1 written solution, award 0 points, for 2 or 3 solutions, award 1 point, for 4 or 5 solutions, award 2 points, and for all solutions, award 3 points)