2A. Determine all pairs of integers $x$ and $y$ such that

$$
1 + 2026x + 2028y = xy
$$
Solution. The given equation is successively equivalent to the equations

$$
\begin{aligned}
& x y-2026 x-2028 y+2026 \cdot 2028=1+2026 \cdot 2028 \\
& x(y-2026)-2028(y-2026)=1+2026(2+2026) \\
& (x-2026)(y-2026)=1+2 \cdot 2026+2026^{2} \\
& (x-2028)(y-2026)=(1+2026)^{2} \\
& (x-2028)(y-2026)=2027^{2}
\end{aligned}
$$

The number 2027 is prime, so from the last equation we get

$$
x-2028 \in\left\{ \pm 1, \pm 2027, \pm 2027^{2}\right\}
$$

from which we obtain the following solutions to the given equation

$$
\begin{array}{r}
(x, y) \in\left\{\left(2029,2027^{2}+1\right),\left(2027,-2027^{2}+2026,(4055,4053)\right.\right. \\
\left.(1,-1),\left(2027^{2}+2028,2027\right),\left(-2027^{2}+2028,2025\right)\right\}
\end{array}
$$