8. (1963 24th Putnam Mathematical Competition) Suppose $x^{2}-x+a$ divides $x^{13}+x+90$, determine the value of the positive integer $a$.

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8. 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.

Let $x=-1,0,1$, then
$$
\left\{\begin{array}{l}
(a+2) q(-1)=88, \\
a q(0)=90, \\
a q(1)=92 .
\end{array}\right.
$$
(2)
(3)
From (2) and (3), we get
$a[q(1)-q(0)]=2$, so $a \mid 2$.
Therefore, $a=1$ or 2.
But when $a=1$, from (1) we have
$3 q(-1)=88$, which is impossible.
Hence, we must have $a=2$.
When $a=2$, we have
$$
x^{13}+x+90=\left(x^{2}-x+2\right)\left(x^{11}+x^{10}-x^{9}-3 x^{8}-x^{7}+5 x^{6}+7 x^{5}-3 x^{4}-17 x^{3}-11 x^{2}+23 x+45\right) .
$$