Problem 9.1. Find all values of the real parameters $a$ and $b$ such that the remainder in the division of the polynomial $x^{4}-3 a x^{3}+a x+b$ by the polynomial $x^{2}-1$ is equal to $\left(a^{2}+1\right) x+3 b^{2}$.

Peter Boyvalenkov
9.1. It follows from the condition that there is a polynomial $q(x)$ of degree 2 such that

$$
x^{4}-3 a x^{3}+a x+b=q(x)\left(x^{2}-1\right)+\left(a^{2}+1\right) x+3 b^{2}
$$

for every $x$. Setting $x=1$ and $x=-1$ we obtain the system

$$
\left\lvert\, \begin{array}{cc}
a^{2}+3 b^{2}+2 a-b=0 \\
a^{2}-3 b^{2}+2 a+b+2=0
\end{array}\right.
$$

and it is easy to see that $a=-1, b=\frac{1 \pm \sqrt{13}}{6}$. A direct verification shows that these values of $a$ and $b$ are solutions indeed.