1A.2B. In the set of integers, solve the equation

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

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

The number 2027 is prime, so $x-2028 \in\left\{ \pm 1, \pm 2027, \pm 2027^{2}\right\}$, from which we get

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