Problem 1290

III/2. Let $P(x)=\alpha x^{2}+\beta x+\gamma$. Then the numbers $P(0)=\gamma, P(1)-P(0)=$ $(\alpha+\beta+\gamma)-\gamma=\alpha+\beta$ and $P(2)-2 P(1)+P(0)=(4 \alpha+2 \beta+\gamma)-2(\alpha+\beta+\gamma)+\gamma=2 \alpha$ must be integers. Let $2 \alpha=a \in \mathbb{Z}$, so $\alpha=\frac{a}{2}$. If $a$ is even, the number $\alpha$ is an integer, so the number $\beta$ must also be an integer, which is in contradiction with the assumption of the problem, since all coefficients of the polynomial $P$ are integers. Therefore, $a$ must be odd, i.e., of the form $a=2 a^{\prime}+1, a^{\prime} \in \mathbb{Z}$. Thus, $\alpha=a^{\prime}+\frac{1}{2}$. From this, it follows that $\beta$ must also be of the form $\beta=b+\frac{1}{2}, b \in \mathbb{Z}$. Therefore, $P(x)=a^{\prime} x^{2}+\frac{1}{2} x^{2}+b x+\frac{1}{2} x+\gamma$, so the polynomial $Q(x)=a^{\prime} x^{2}+b x+\gamma$ has only integer coefficients.

2nd method. Let $P(x)=\alpha x^{2}+\beta x+\gamma$. Then the numbers $P(0)=\gamma, P(1)-P(0)=$ $(\alpha+\beta+\gamma)-\gamma=\alpha+\beta$ and $P(-1)-P(0)=(\alpha-\beta+\gamma)-\gamma=\alpha-\beta$ must be integers. Thus, we know that
$\alpha+\beta, \alpha-\beta \in \mathbb{Z}$. If we add and subtract the last two numbers, we find that the numbers $2 \alpha$ and $2 \beta$ must also be integers. If $2 \alpha$ is an even number, $\alpha$ is an integer. Since $\alpha+\beta \in \mathbb{Z}$, $\beta$ is also an integer - the condition of the problem is therefore not met. Thus, $2 \alpha$ must be an odd number. By a similar argument, we find that $2 \beta$ must also be an odd number.

Thus, we can write $2 \alpha=2 a+1, a \in \mathbb{Z}$ and $2 \beta=2 b+1, b \in \mathbb{Z}$. We substitute these into the polynomial $Q$ and get $Q(x)=a x^{2}+b x+\gamma$. All its coefficients are indeed integers.
Conclusion $\gamma \in \mathbb{Z}$ ..... 1 point
Conclusion $\alpha+\beta \in \mathbb{Z}$ ..... 1 point
Conclusion $2 \alpha$ or $2 \beta \in \mathbb{Z}$ ..... 1 point
Use of the assumption and conclusion that two coefficients are not integers ..... 1 point
Conclusion that $\alpha$ is of the form $a+\frac{1}{2}$ for some $a \in \mathbb{Z}$ ..... 1 point
Conclusion that $\beta$ is of the form $b+\frac{1}{2}$ for some $b \in \mathbb{Z}$ ..... 1 point
Final conclusion that the coefficients of the polynomial $Q$ are integers ..... 1 point

The last point is awarded only for completely irrefutable reasoning throughout the ENTIRE solution. If the contestant assumes from the start that the coefficients are rational numbers (and immediately writes the coefficients in the form $\frac{x}{y}$), the conclusions drawn by this method are NOT RECOGNIZED.