Find all integers $a$ for which $x^2-x+a$ divide $x^{13}+x+90$. Find all integers $a$ for which $x^2-x+a$ divide $x^{13}+x+90$.
The answer is $a=2$. 
 A: If $a$ is negative or zero, then the quadratic has two real roots.
But we can easily check that the other polynomial has derivative everywhere positive
and hence only one real root.
So $a$ must be positive. 
If $x^2-x+a$ divides $x^{13}+x+90$, then $x^{13}+x+90=f(x)(x^2-x+a)$,
where $f(x)$ is a polynomial with integer coefficients.
Let $x=0$, we see that $a$ must divide $90$. Let $x=1$, we see that it must divide $92.$
Hence it must divide $92-90=2.$ So the only possibilities are $1$ and $2.$
Suppose $a=1$, then putting $x=2$, we have that $3$ divides $2^{13}+92$ but $2^{\text{odd}}$ is congruent to $2 \mod 3,$ so $2^{13}+92$ is congruent to $1 \mod 3.$ 
So $a$ cannot be $1.$ 
To see that $a=2$ is possible, we write
$(x^2-x+2)(x^{11}+x^{10}-x^9-3x^8-x^7+5x^6+7x^5-3x^4-17x^3-11x^2+23x+45)=x^{13}+x+90$.
A: Hint : You only have to check the divisors of $90$, but do not forget the negative ones.
A: Relocating my answer from a new incarnation of the same question. I am using a somewhat more complicated way of eliminating the larger choices of $a$ :-(

The high degree polynomial has a single real zero, so the quadratic cannot have any, and thus $a$ must be a positive factor of $90$. We get a useful bound on $a$ by observing that the (complex) zeros of $x^{13}+x+90$ must have absolute value $\le 3/2$. For if $|z|\ge3/2$, then the triangle inequality tells that 
$$|z^{13}+z+90|\ge (3/2)^{13}-3/2-90>100.$$
But the zeros of $x^2-x+a$ are $x=(1\pm\sqrt{1-4a})/2$. They have squared absolute value
$$|x|^2=\frac{1+(4a-1)}{4}=a.$$
This leaves $a=2$ and $a=1$ as the only alternatives.
If $a=1$, the roots of $x^2-x+1$ are the sixth roots of unity with absolute value $=1$ – too small to be zeros of $x^{13}+x+90$.
$a=2$ works. Verifying this is a bit tedious. The roots are $z_{1,2}=(1+i\sqrt{7})/2$. Their powers satisfy the recurrence
$$
z^{n+2}=z^{n+1}-2z^n,
$$
and running this a dozen iterations gives the claim. An alternative may be to observe that they are the eigenvalues of the companion matrix
$$
A=\pmatrix{0&1\cr-2&1\cr}.
$$
It is quick to calculate $A^{13}$ by the good ol' square-and-multiply, and the answer is that
$$A^{13}+A=-90I$$ 
as required.
