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

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The discussion revolves around finding all integer values of \( k \) such that the polynomial \( x^2 - x + k \) divides \( x^{13} + x + 90 \). Participants express a preference for the polynomial to be \( x^2 - x - k \) instead. Clarification is provided that \( k \) is an integer, not limited to natural numbers. The thread hints at the complexity of the problem and the need for analytical methods to derive the solution. Overall, the focus remains on determining the specific integer values of \( k \) that satisfy the divisibility condition.
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Find all integers $k$ for which $x^2-x+k$ divides $x^{13}+x+90$.
 
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anemone said:
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 [math]x^2 - x - k[/math]...

-Dan

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

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

$(x^2-x+2)(x^{11}+x^{10}-x^9-3x^8-x^7+5x^6+7x^5-3x^4-17x^3-11x^2+23x+45)=x^{13}+x+90$.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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