2. (2 points) For which integers $p$ and $q$ is the value

of the polynomial $f(x)=x^{2}+p x+q$ divisible by 2 for any integer $x ?$
Solution: According to the condition

$f(x+1)-f(x)=2 x+1+p \vdots 2 \forall x \in Z \Leftrightarrow(p+1) \vdots 2 \Leftrightarrow p-$ is odd. Therefore, the numbers $x^{2}$ and $p x$ are either both even or both odd, which means $x^{2}+p x \vdots 2$. Consequently, $q$ is even.

Answer: $p$ is odd, $q$ is even.