Find the complex $a$,such $x^2-x+a|x^{13}+x+90$

Question

If $a$ be complex number,such $$x^2-x+a|x^{13}+x+90$$find $a$

I think it is equality if $r^2-r+a=0$ $$\Longrightarrow r^{13}+r+90=0$$,find the $a$

Answer 1

The following verifies the uniqueness of the solution $a=2$ proposed in previous answers, by brute force (and some WA-assistance)... Let $x_{1,2}$ be the roots of $x^2-x+a$, then by Vieta's relations $x_1+x_2=1$ and $x_1x_2=a$. For divisibility, they must also be roots of $x^{13}+x+90$, so:

$$ x_1^{13}=-(x_1+90) \\ x_2^{13}=-(x_2+90) $$

Multiplying the above and simplifying:

$$ x_1^{13}x_2^{13}=x_1x_2 +90(x_1+x_2)+90^2 \quad\iff\quad a^{13}-a-8190 = 0 \tag{1} $$

It can be found that $(1)$ has the root $a=2$ by the rational root theorem (or by previously proposed heuristics), and it can be verified that it satisfies the given problem since:

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

Factoring $a-2$ out of $(1)$ leaves:

$$ a^{12} + 2 a^{11} + 4 a^{10} + 8 a^9 + 16 a^8 + 32 a^7 \\+ 64 a^6 + 128 a^5 + 256 a^4 + 512 a^3 + 1024 a^2 + 2048 a + 4095 = 0 \tag{2} $$

In order to conclude that $a=2$ is the unique solution, what's left to show is that none of the $\,a\,$roots of $(2)$ corresponds to $\,x\,$ roots of $x^2-x+a=0$ which also satisfy $x^{13}+x+90=0$.

Indeed, eliminating $a$ between $(2)$ and $x^2-x+a=0$ by way of polynomial resultants gives a $24^{th}$ degree polynomial in $x$:

$$ x^{24} - 12 x^{23} + 64 x^{22} - 198 x^{21} + 389 x^{20} - 502 x^{19} + 436 x^{18} - 276 x^{17} \\ + 139 x^{16} - 24 x^{15} - 16 x^{14} - 46 x^{13} - x^{12} + 90 x^{11} + 92 x^{10} - 88 x^9 \\ - 272 x^8 - 96 x^7 + 448 x^6 + 640 x^5 - 256 x^4 - 1536 x^3 - 1024 x^2 + 2048 x + 4095 $$

It can be verified that the above has no common factor with $x^{13}+x+90$ by Euclidean division.

Answer 2

$$x^2-x+a|x^{13}+x+90$$ Hoping you may get lucky finding an 'integer' $a$, here is a simple shortcut
When $x=0$, you want $a | 90$
When $x=1$, you want $a | 92$

Can you guess $a$ that satisfies both above requirements

Answer 3

As remarked by @rsadhvika, $a=\pm 1 $ or $\pm 2$. If $a=-2$ then the polynomial has $-1$ as a root, this gives a contradiction. If $a=+1$ let $x=2$ this gives a non divisiblity, and if $a=-1$, let $x=-2$, another non divisibility. So $a=2$.

Answer 4

Is cheating allowed?

>> P=[1 0 0 0 0 0 0 0 0 0 0 0 1 90]

P =

     1     0     0     0     0     0     0     0     0     0     0     0     1    90

>> r=roots(P)

r =

   1.3740 + 0.3391i
   1.3740 - 0.3391i
   1.0583 + 0.9391i
   1.0583 - 0.9391i
   0.5000 + 1.3229i
   0.5000 - 1.3229i
  -0.1721 + 1.4029i
  -0.1721 - 1.4029i
  -0.8036 + 1.1618i
  -0.8036 - 1.1618i
  -1.4119 + 0.0000i
  -1.2507 + 0.6555i
  -1.2507 - 0.6555i

>> r(5)*r(6)

ans =

    2.0000

So we found all the roots of the polynomial and selected the $2$ roots that added to $1$ so as to match the linear term of the quadratic factor. Then their product was $a=2$.