Estonian Mathematical Olympiad

Suppose that the first player always decreases the constant term by the smaller solution. We show that this is a winning strategy. Assume the opposite, i.e., that the play lasts infinitely. As the combined effect of the first and the second player's moves, the coefficient at the linear term either increases by 1 or (if the second player uses the alternative move) remains the same. If the second player had used the main move infinitely many times, the coefficient at the linear term would have become positive sooner or later. By Viéte's theorem, the sum of the solutions would be negative. This means that at least one solution is negative, too,

whence the first player must have won already. Consequently, the second player must have used only the alternative move starting from some position on. If the equation on the blackboard before the first player's move is $x^2+px+q=0$ with solutions $x_1,x_2$, where $x_1\le x_2$, then Viéte's theorem implies $p=-(x_1+x_2)$ and $q=x_1{x}_2$, whence after the first player's move the coefficient at the linear term is $-(x_1+x_2)+1$ and the constant term is $x_1{x}_2-x_1$. This means that the quadratic equation after the first player's move is $x^2-(x_1+(x_2-1))+x_1(x_2-1)$. By Viéte's theorem, this equation has solutions $x_1$ and $x_2-1$. Note that if before the first player's move the solutions of the equations differ by more than 1, then the difference decreases by 1 as the result of the move, but if the solutions before the move differ by at most 1 then the difference of the solutions after the move is still at most 1. Analogously, the same holds for the second player's alternative move. Hence the play must reach a position where the second player cannot make the alternative move, contradiction.