Problem 10.5. Integers $a$ and $b$ are such that the quadratic trinomials $x^{2}+a x+b$ and $x^{2}+$ $b x+1100$ have a common root, which is a prime number. Find $a$. List all possible options.
Answer: 40 and 274.

Solution. Let $p$ be the common root of the quadratic trinomials, which is a prime number. Then $p^{2}+a p+b=0$, from which we can conclude that $b \vdots p$.

Now let's look at the second quadratic trinomial: $p^{2}+b p+1100=0$. From the fact that $b \vdots p$, we conclude that $1100=2^{2} \cdot 5^{2} \cdot 11 \vdots p^{2}$, that is, $p=2$ or $p=5$.

In either case, $b$ can be expressed from the equality $p^{2}+b p+1100=0$ (resulting in -552 and -225 respectively); then $a$ is similarly expressed from the equality $p^{2}+a p+b=0$ - resulting in 274 and 40 respectively.