1. The quadratic trinomial $x^{2}+p x+q$ has two distinct non-zero integer roots $a$ and $b$. It is known that $a+p$ is divisible by $q-2 b$. What can the root $a$ be equal to? (Provide all answers and prove that there are no others.)
Answer: $a=3$ or $a=1$.

Solution. By Vieta's theorem, $p=-a-b, q=ab$. Then, according to the condition, the number $a+p=a+(-a-b)=-b$ is divisible by $q-2b=ab-2b=(a-2)b$. When $a=2$, the last number equals zero, and it is incorrect to talk about divisibility. When $a=3$ or $a=1$, divisibility holds. For other integer values of $a$, divisibility is not possible, as the divisor in absolute value will be greater than the dividend. Note that both found answers are suitable, for example, for the polynomial $x^{2}-4x+3$.

Comment. Vieta's theorem is written, and the problem is reduced to the divisibility of (-b) by (a-2)b - 4 points. It is proven that divisibility holds only for two values of a - 3 points. Points are summed.