Let the polynomial formed be $P(x) = a_{n} x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \dots + a_1 x + a_0$. For the first part of the problem, we will prove that the second player can guarantee there exists an integer root of $P(x)$ if $n$ is odd. Group the coefficients as follow: \[ \{ a_n , a_{n-1} \} ; \{ a_{n-2} , a_{n-3} \} ; \dots ; \{ a_3, a_2 \}, \{a_1 , a_0 \} \] Now, every time the first player moves, as an example, $a_k = x$. Choose the one with the same set as $a_k$ and change it to $x$. We'll prove that this strategy works. Note that we can now group the polynomial $P(x) = a_n x^n + a_n x^{n-1} + a_{n-2} x^{n-2} + a_{n-2} x^{n-1} + \dots + a_3 x^3 + a_3x^2 + a_1 x + a_1 $ It is obvious to see that $-1$ is indeed a root.