jmc

We have that $$(x^2-x+a)p(x)=x^{13}+x+90$$for some polynomial $p(x)$ with integer coefficients.

Setting $x=0,$ we get $ap(0)=90.$ This means $a$ divides 90.

Setting $x=1,$ we get $ap(1)=92.$ This means $a$ divides 92.

Since $a$ divides both 90 and 92, it must divide $92-90=2.$ Hence, $a$ must be equal to 2, 1, $-1,$ or $-2.$

Setting $x=-1,$ we get $(a+2)p(-1)=88.$ This means $a+2$ divides 88. Of the four values we listed above, only $a=-1$ and $a=2$ work.

If $a=-1,$ then $x^2-x+a$ becomes $x^2-x-1=0$. The roots are $$x=\frac{1\pm \sqrt{5}}{2}.$$In particular, one root is positive, and one root is negative. But $x^{13}+x+90$ is positive for all positive $x,$ which means that it does not have any positive roots. Therefore, $a$ cannot be $-1,$ which means $a=\boxed{2}.$

By Long Division, $$x^{13}+x+90=(x^2-x+2)(x^{11}+x^{10}-x^9-3x^8-x^7+5x^6+7x^5-3x^4-17x^3-11x^2+23x+45).$$