Example 3: Let $a$ be an integer such that $x^{2}-x+a$ divides $x^{13}+x+90$. Find the value of $a$.

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Analysis: This problem can be solved by dividing $x^{13}+x+90$ by $x^{2}-x+a$, and setting the coefficients of the resulting remainder (at most a linear term) to 0 to find $a$. Alternatively, the roots of $x^{2}-x+a=0$, $x=\frac{1 \pm \sqrt{1-4 a}}{2}$, can be substituted into $x^{13}+x+90=0$ to find $a$. However, both methods are relatively cumbersome.

From the problem statement, we know that there must be an integer-coefficient polynomial $q(x)$ such that $x^{13}+x+90=\left(x^{2}-x+a\right) q(x)$. This is an identity that holds for all values of $x$. Therefore, by analyzing the values of $x=0, \pm 1$, we can obtain the following solution.

Solution: From the problem statement, we know that there must be an integer-coefficient polynomial $q(x)$ such that $x^{13}+x+90=\left(x^{2}-x+a\right) q(x)$. Note that the equation $x^{13}+x+90=0$ clearly has no positive roots, so $x^{2}-x+a=0$ must also have no positive roots. The roots of $x^{2}-x+a=0$ can be expressed as $x=\frac{1 \pm \sqrt{1-4 a}}{2}$. This implies $1-4 a < 0$.
Let $x=0$, from (1) we know $a \mid 90$.
Let $x=1$, from (1) we know $a \mid 92$. Therefore, $a \mid (92-90)$, i.e., $a \mid 2$. Given $a>0$, we have $a=1$ or $a=2$. Let $x=-1$, from (1) we know $(a+2) \mid 88$. Therefore, $a=1$ is not possible.
In summary, $a=2$.