Find the 'a' such that $x^{13}+x+90=(x^2-x+a)(Q(x))$

Question

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.

Answer 1

Substituting $x = 1 \implies a \mid 92$

Substituting $x = -1 \implies (a+2)\mid 88$

Divisors of $92 = 1,2,4,23,46,92$

Divisors of $88 = 1,2,4,8,11,22,44,88$

Thus, $a = 2$.

Indeed, checks out with Wolfram :

Answer 2

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 $a=2$.

Answer 3

A remark and an alternative proof.

Remark: this problem as already been posed in the 1963 Putnam competition

https://mks.mff.cuni.cz/kalva/putnam/psoln/psol637.html

An alternative proof:

The arithmetical proof of @Ojas is perfect, but another one is possible.

I was in fact puzzled by the fact that when I solved (with the aid of Mathematica) the long division by $x^2-x+a$ ($a$ unknown), I obtained the same solution as you, @Ojas,without assuming integer (or rational) values for the coefficients; an afterthought said me that it is perfectly normal because long division with a single unknown coefficient gives (at most) a solution.

Then I thought that there should exist a non-arithmetical proof. Here is one.

Let $P(x)=x^{13}+x+90$; having an odd degree and being strictly increasing (its derivative is everywhere $>0$), $P$ has a unique real root. Thus its factorisation into irreducible factors in $\mathbb{R}[x]$ is a product of a first degree polynomial and 6 second degree polynomials, the latter having complex conjugate roots of $P$.

As a second degree factor $F(x)=x^2-x+a$ is imposed in the question, and because there is only one real root, this real root is a root of $Q(x)$. As the sum of the roots of $F$ is $1$, we have to look for conjugate roots of the form $\frac12 \pm bi$. There are such roots: $r_1, r_2=\frac12(1 \pm i\sqrt{7})$, whence $a=r_1r_2=\frac84=2$, proving the result.

Frankly speaking, I admit that $\frac12(1 \pm i\sqrt{7})$ is the rabbit pulled out of the hat...

By mere curiosity, I have represented on the same figure the roots $z_k$ of $P$ by blue dots and the roots of the "perturbated" polynomial $z^{13}+90$ i.e., $\zeta_k=re^{i(2k+1)\pi/13} \ (k \in \mathbb{Z})$ by black stars, with $r=\sqrt[13]{90}$. The two families are rather close. Roots $z_3$ and $z_{11}$ correspond to $r_1$ and $r_2$ and $z_7$ is the real root.