For which integer $a$ does $x^2 - x + a$ divide $x^{13} + x + 90$?
Let's try to divide $x^{13} + x + 90$ by $x^2 - x + a$. We should get a remainder of 0 for this to work. Yeah, so let's do polynomial long division. Ok, so the first term of our dividend is $x^{13}$, and the first term of our divisor is $x^2$. So $x^{13}$ divided by $x^2$ is $x^{11}$. And $x^{11}$ times $x^2$ is $x^{13}$. So our first term matches. The next term of the dividend is $x$, and the next term of the divisor is $-x$. And since $x^{12}$ divided by $x^2$ is $x^{10}$, the next term of the quotient is $x^{10}$. So $x^{10}$ times $x^2$ is $x^{12}$. And $x^{10}$ times $-x$ is $-x^{11}$. Right. So $a$ must be a factor of 90 and 92. The factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90. The factors of 92 are 1, 2, 4, 23, 46, and 92. So $a$ can be $1$ or $2$. Let's see if either of these work. If $a=1$, then our divisor is $x^2-x+1$, and our dividend is $x^{13}+x+90$. That's $-1-1+90=88$ for the dividend, and $1+1+1=3$ for the divisor. 88 is not divisible by 3, so $a=1$ doesn't work. Let's try $a=2$. Our divisor is $x^2-x+2$, and our dividend is $x^{13}+x+90$. So the remainder is 0, and $a=2$ works. So the only integer $a$ that works is $2$.

