Slovenia — National Math Olympiad, 2012

Denote $P(x) = \alpha x^2 + \beta x + \gamma$. Then $P(0) = \gamma$, $P(1) - P(0) = (\alpha + \beta + \gamma) - \gamma = \alpha + \beta$ and $P(-1) - P(0) = (\alpha - \beta + \gamma) - \gamma = \alpha - \beta$ must be integers. We thus know that $\alpha + \beta, \alpha - \beta \in \mathbb{Z}$. If we add and subtract the last two numbers, we notice that $2\alpha$ and $2\beta$ must also be integers. If $2\alpha$ is even, then $\alpha$ is an integer. Because $\alpha + \beta \in \mathbb{Z}$, $\beta$ is also an integer, and the condition of the problem is not fulfilled. From this we conclude that $2\alpha$ is odd. A similar reasoning gives that $2\beta$ is also odd.

We can thus write $2\alpha = 2a + 1$, $a \in \mathbb{Z}$ and $2\beta = 2b + 1$, $b \in \mathbb{Z}$. If we substitute these into the polynomial $Q$, we get $Q(x) = a x^2 + b x + \gamma$. All its coefficients are indeed integers.