
09.2. On a faded piece of paper it is possible, with some effort, to discern the following:

$$
\left(x^{2}+x+a\right)\left(x^{15}-\ldots\right)=x^{17}+x^{13}+x^{5}-90 x^{4}+x-90
$$

Some parts have got lost, partly the constant term of the first factor of the left side, partly the main part of the other factor. It would be possible to restore the polynomial forming the other factor, but we restrict ourselves to asking the question: What is the value of the constant term a? We assume that all polynomials in the statement above have only integer coefficients.


Solution. We denote the polynomial $x^{2}+x+a$ by $P_{a}(x)$, the polynomial forming the other factor of the left side by $Q(x)$ and the polynomial on the right side by $R(x)$. The polynomials are integer valued for every integer $x$. For $x=0$ we get $P_{a}(0)=a$ and $R(0)=-90$, so $a$ is a divisor of $90=2 \cdot 3 \cdot 3 \cdot 5$. For $x=-1$ we get $P_{a}(-1)=-184$, so $a$ is also a divisor of $184=2 \cdot 2 \cdot 2 \cdot 23$. But the only prime factor in common is 2 . So the only possibilities for $a$ are $\pm 2$ and $\pm 1$. If $a=1$, we get for $x=1$ that $P_{1}(1)=3$, while $R(1)=4-180=-176$, which cannot be divided by 3 . If $a=-2$ we get for $x=1$ that $P_{2}(1)=0$, i.e. the left side is equal to 0 , while the right side is equal to $R(1)=-176$, which is different from 0 . Neither $a=1$ nor $a=-2$ will thus work. It remains to check $a=2$ and $a=-1$. Before we use the procedure above again, we need a factorization of $R(x)$. We observe that $x^{4}+1$ is a divisor of $R(x)$, since the right side may be written as $\left(x^{4}+1\right)\left(x^{13}+x-90\right)$. If $a=-1$ we get for $x=2$ that $P_{1}(2)=5$, while $x^{4}+1=17$ and $x^{13}+x-90=8104$. So the right hand side is not divisible by 5 . Now, the only remaining possibility is $a=2$, i.e. $x^{2}+x+2$ is a divisor of $R(x)$.
