\section*{Problem 3B - 321043B}

Determine all integers \(a, b\) for which the division of the polynomial

\[
f(x)=x^{4}+2 a x^{3}+2 x^{2}+x-2
\]

by the polynomial

\[
g(x)=x^{2}+a x+b
\]

results in a polynomial \(h(x)\) with no remainder (such that for every number \(x\) for which \(g(x) \neq 0\), the equation \(f(x): g(x)=h(x)\) holds).
}

It is

\[
\begin{gathered}
f(x)-g(x) \cdot\left(x^{2}+a x\right)=\left(x^{4}+2 a x^{3}+2 x^{2}+x-2\right)-\left(x^{4}+a x^{3}+b x^{2}+a x^{3}+a^{2} x^{2}+a b x\right)= \\
=\left(2-b-a^{2}\right) x^{2}+(1-a b) x-2 \quad \text { and } \\
g(x) \cdot\left(2-b-a^{2}\right)=\left(2-b-a^{2}\right) x^{2}+\left(2 a-a b-a^{3}\right) x+\left(2 b-b^{2}-a^{2} b\right)
\end{gathered}
\]

thus with \(h(x):=x^{2}+a x+2-b-a^{2}\) :

\[
\begin{gathered}
f(x)-g(x) \cdot h(x)=\left(2-b-a^{2}\right) x^{2}+(1-a b) x-2-\left(\left(2-b-a^{2}\right) x^{2}+\left(2 a-a b-a^{3}\right) x+\left(2 b-b^{2}-a^{2} b\right)\right)= \\
\quad=\left(1-a b-2 a+a b+a^{3}\right) x-2-2 b+b^{2}+a^{2} b=\left(1-2 a+a^{3}\right) x-2-2 b+b^{2}+a^{2} b
\end{gathered}
\]

For no \(x\) to leave a remainder, both \(1-2 a+a^{3}=0\) and \(2-2 b+b^{2}+a^{2} b=0\) must hold.

First, consider the equation \(1-2 a+a^{3}=0\). This has the solution \(a_{1}=1\), so we can divide the equation by \(a-1\) and obtain \(0=a^{2}+a-1\), which has the further solutions \(a_{2 / 3}=\frac{-1 \pm \sqrt{5}}{2} \notin \mathbb{Z}\). Therefore, the only integer solution is \(a=1\).

Substituting this into the second equation, we must have \(2-2 b+b^{2}+b=0\) or \((-2+b)(1+b)=0\), so \(b \in\{-1,2\}\).

Thus, there are two integer pairs \((a ; b)\) such that the polynomial division of \(f\) by \(g\) leaves no remainder, namely \((1 ;-1)\) and \((1 ; 2)\). The verification confirms these solutions.