$1 \cdot 47$ Let $x^{2}-x+a$ divide $x^{13}+x+90$, determine the positive integer value of $a$.
(24th Putnam Mathematical Competition, 1963)
[Solution] Let $x^{13}+x+90=\left(x^{2}-x+a\right) q(x)$, where $a$ is an integer, and $q(x)$ is a polynomial with integer coefficients.
$$
\begin{array}{c}
\text { Let } x=-1,0,1, \text { then } \\
\left\{\begin{array}{l}
(a+2) q(-1)=88, \\
a q(0)=90, \\
a q(1)=92 .
\end{array}\right.
\end{array}
$$

From (2),(3),
$$
a[q(1)-q(0)]=2,
$$

Thus,
$$
a \mid 2 \text {. }
$$

Therefore,
$$
a=1 \text { or } 2 \text {. }
$$

But when $a=1$, from (1) we get
$$
3 q(-1)=88 \text {. }
$$

This is impossible.
Hence, we can only have $a=2$.
When $a=2$, we have
$$
\begin{aligned}
x^{13}+x+90= & \left(x^{2}-x+2\right)\left(x^{11}+x^{10}-x^{9}-3 x^{8}-x^{7}+5 x^{6}+\right. \\
& \left.7 x^{5}-3 x^{4}-17 x^{3}-11 x^{2}+23 x+45\right) .
\end{aligned}
$$