Question 41
Maryam and Artur play a game on a board, taking turns. At the beginning, the polynomial $XY - 1$ is written on the board. Artur is the first to make a move. In each move, the player replaces the polynomial $P(X, Y)$ on the board with one of the following polynomials of their choice:
a) $X \cdot P(X, Y)$
b) $Y \cdot P(X, Y)$
c) $P(X, Y) + a$, where $a \in (-\infty, 2025]$ is an arbitrary integer.
The game stops after both players have made 2025 moves. Let $Q(X, Y)$ be the polynomial on the board after the game ends. Maryam wins if the equation $Q(x, y) = 0$ has a finite and odd number of positive integer solutions $(x, y)$. Prove that Maryam can always win the game, no matter how Artur plays.
a) $X \cdot P(X, Y)$
b) $Y \cdot P(X, Y)$
c) $P(X, Y) + a$, where $a \in (-\infty, 2025]$ is an arbitrary integer.
The game stops after both players have made 2025 moves. Let $Q(X, Y)$ be the polynomial on the board after the game ends. Maryam wins if the equation $Q(x, y) = 0$ has a finite and odd number of positive integer solutions $(x, y)$. Prove that Maryam can always win the game, no matter how Artur plays.