## Task A-1.1.

Determine all ordered pairs of integers $(x, y)$ for which

$$
x^{2} y+4 x^{2}-3 y=51
$$
## Solution.

Let's rewrite the initial equality in a different way:

$$
\begin{aligned}
x^{2} y+4 x^{2}-3 y & =51 & & \\
x^{2}(y+4)-3 y & =51 & & \\
x^{2}(y+4)-3(y+4)+12 & =51 & & 1 \text { point } \\
\left(x^{2}-3\right)(y+4) & =39 . & & 2 \text { points }
\end{aligned}
$$

The factors \(x^{2}-3\) and \(y+4\) on the left side are divisors of the number 39, and their product is 39. We conclude that

$$
x^{2}-3 \in\{-39,-13,-3,-1,1,3,13,39\}
$$

which means \(x^{2} \in\{-36,-10,0,2,4,6,16,42\}\). Of these numbers, only 0, 4, and 16 are perfect squares.

For \(y\), we have \(y+4=\frac{39}{x^{2}-3}\).

If \(x^{2}=0\), i.e., \(x=0\), then \(y=-17\). 1 point

If \(x^{2}=4\), i.e., \(x= \pm 2\), then \(y=35\). 1 point

If \(x^{2}=16\), i.e., \(x= \pm 4\), then \(y=-1\).

We conclude that all solutions to the initial equation are

$$
(x, y)=(0,-17),(2,35),(-2,35),(4,-1),(-4,-1)
$$