Find Integer $k$: x^2-x+k Divides x^13+x+90

[anemone]

Find all integers $k$ for which $x^2-x+k$ divides $x^{13}+x+90$.

 

[topsquark]

Find all integers $k$ for which $x^2-x+k$ divides $x^{13}+x+90$.

Analytically speaking I'd love you more if that were \(\displaystyle x^2 - x - k\)...

-Dan

Edit: Oh k is an integer, not a natural number. Okay, (Hug)

 

[anemone]

Solution of other:

Spoiler

If $k$ 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 $k$ must be positive.

If $x^2-x+k$ divides $x^{13}+x+90$, then $x^{13}+x+90=f(x)(x^2-x+k)$, where $f(x)$ is a polynomial with integer coefficients.

Let $x=0$, we see that $k$ 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 $k=1$, 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 $k$ cannot be 1.

To see that $k=2$ is possible, we write

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