, 2025

We claim that Maryam can always achieve that the polynomial on the board at the end of her turn has the form $P(X,Y) = f(XY)$ where $f \in \mathbb{Z}[T]$ can be written as
$$T^n - \sum_{i=0}^{n-1} a_i T^i \quad \text{for integers } n > 0 \text{ and } a_i \ge 0, \text{ not all of them zero.} \quad (1)$$

As any such $f$ fulfills $\frac{f(x)}{x^n} = 1 - \sum_{i=0}^{n-1} \frac{a_i}{x^{n-i}}$ for all positive real numbers $x$, which is a strictly increasing function on $(0, \infty)$ with arbitrarily small real values near $0$ and tending to $1$ for $x \to \infty$, it has exactly one positive real root $r$. We claim further that Maryam can choose $r$ to be a perfect (integer) square. (Initially, $P(X, Y) = f(XY)$ with $f = T - 1$.) Maryam proceeds as follows:

* If Artur multiplies with $X$, Maryam multiplies with $Y$ and if Artur multiplies with $Y$, Maryam multiplies with $X$. If initially, $P(X, Y) = f(XY)$ was on the board, then the resulting polynomial is $XYf(XY)$, so $f$ changes to $T \cdot f$.

* If Artur adds an integer $0 \le a \le 2025$, Maryam adds the number $-a \le 2025$. The polynomial remains unchanged.

* If Artur adds an integer $a < 0$, write $A(XY)$ for the new polynomial on the board, where $A \in \mathbb{Z}[T]$ is of the form (1). By the discussion above, $A$ has a unique positive real root $u$. As $A(x) > 0$ for all $x > u$, Maryam can choose an integer $c > u$ (e.g. $c = \lfloor u \rfloor + 1$) and add the negative integer $-A(c^2)$ to the polynomial $A$ on the board. Then the new polynomial on the board has again the form (1) and (by construction) $c^2 \in \mathbb{Z}_{>0}$ as the only positive real root.

Hence, Maryam can always achieve that $Q$ (the polynomial in the end of the game) satisfies $Q = g(XY)$ where $g$ is of the form (1) and has a perfect square $s^2$, $s \in \mathbb{Z}_{>0}$, as unique positive real root. Now for all pairs of positive integers $(x, y)$, we have $Q(x, y) = 0 \iff g(xy) = 0 \iff xy = s^2$ and it is well known that the number of solutions $(x, y)$ to the last equation is (finite and) odd. (Pairs $(x, y)$ and $(y, x)$ with $x \ne y$ correspond and $(s, s)$ is the only fixed point in this involution, giving an odd number overall.) Hence, Maryam can always win, independent of Artur's moves.