26*. Does there exist an integer $a$ such that the polynomial $f(x)=x^{13}+x+90$ is divisible by the trinomial $x^{2}-x+a$?
$\triangle$ Since the polynomial $f(x)$ is divisible by $x^{2}-x+a$, then for any integer $x$ the number $f(x)$ is divisible by the number $x^{2}-x+a$ (in the sense of divisibility of integers):

$$
\left(x^{13}+x+90\right) \vdots\left(x^{2}-x+a\right)
$$

Let's substitute $x=1$, and then $x=0$:

$$
92 \vdots a, 90 \vdots a
$$

From the divisibility of the numbers 92 and 90 by $a$, it follows that 2 :a. Therefore, $a= \pm 1, \pm 2$.

Next, we need to consider four cases depending on $\boldsymbol{a}$. Complete the solution on your own.

Answer: there exists and is unique $-a=2$.