TITLE: Find the 'a' such that $x^{13}+x+90=(x^2-x+a)(Q(x))$
QUESTION [5 upvotes]: Problem: For what integer $a$ does $x^2-x+a$ divide $x^{13}+x+90=P(x)$?
I am struck at this one. I can't think of a good method (using little computations) to do this.
Of course, we could divide $P(x)$ by $x^2-x+a$ and equate the remainder to $0$ or factor out $x^2-x+a$ and use the remainder theorem (But this may not work as we would get an equation with surds and all and 13 degree terms.)
I also tried with the following form:
$x^{13}+x+90=(x^2-x+a)Q(x)$
By differentiating in two different ways I got the following,
$13x^2+1=(2x-1)Q(x)+(x^2-x+a)Q'(x)$
$\frac{(13x^2+1)(x^2-x+a)-(2x-1)(x^{13}+x+90)}{(x^2-x+a)^2}=Q'(x)$
I think we can solve for $Q(x)$ from the last two equations and finish the problem(or do we get an identity?, I haven't tried as it is becoming very cumbersome). 
So, can anyone help me find an elegant solution (will less computations)? 
Thanks.

REPLY [4 votes]: Inspired by Ojas's answer (which I think is incomplete because it seems to assume $a$ is positive):
$x=0\implies a\mid90$
$x=1\implies a\mid92$
Thus $a\mid2$, so $a\in\{\pm1,\pm2\}$
$x=-1\implies(a+2)\mid88\implies a\in\{-1,2\}$
Finally, we can rule out $a=-1$ because $x^2-x-1$ has a root in $[-1,0]$ while $x^{13}+x+90$ clearly does not.  Thus $a=2$ is the only remaining possibility.
Remark:  This doesn't prove that $x^2-x+2$ necessarily is a factor of $x^{13}+x+90$.  All it proves is the following:  If $a$ is an integer such that $x^2-x+a$ divides $x^{13}+x+90$, then $a=2$.