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My question essentially boils down to this: (it was part of a question about polynomial ring ideals)

Find an integer $b$ such that the rational function \begin{equation} \frac{x^5-bx}{x^2-2x}\in \mathbb{Z}[x] \end{equation} I've tried plugging in a bunch of values for $b$ using a calculator but none of them have come out to whole polynomials. Is there an easier way to do this?

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    $\begingroup$ The numerator must be divisible by $x$ so that $a=0$ must hold. Similarly it must be divisible by $x-2$ too because otherwise it cannot lie in the ring $\endgroup$ – Ninja Dec 5 '20 at 22:57
  • $\begingroup$ @Ninja yeah I figured, I just can't find a $b$ that gives the result... $\endgroup$ – Vladimir Lenin Dec 5 '20 at 23:01
  • $\begingroup$ I might edit the question to account for that $\endgroup$ – Vladimir Lenin Dec 5 '20 at 23:01
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    $\begingroup$ In a polynomial ring $R[X]$, if a polynomial $P(X)$ is divisible by $X-r$ for some $r \in R$ then $P(r)=0$ must hold and in your case $x^4-b$ must be divisible by $x-2$. $\endgroup$ – Ninja Dec 5 '20 at 23:05
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You may be guessing a long time at that rate! Write $$ x^5 - bx - a = (x^2-2x)(-------), $$ and fill in the divisor's bracket term by term. To multiply to $x^5$, the first term needs to be $x^3$: $$ x^5 - bx - a = (x^2-2x)(x^3-----). $$ Now $2x^4$ is being subtracted, so you better add $2x^2$ to the divisor to make that up. $$ x^5 - bx - a = (x^2-2x)(x^3+2x^2----). $$ Then $$ x^5 - bx - a = (x^2-2x)(x^3+2x^2+4x--). $$ So then $$ x^5 - bx - a = (x^2-2x)(x^3+2x^2+4x+8), $$ and you can read off $b = 16$, $a = 0$.

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If $x^2-2x=x(x-2)$ divides $x^5-bx=x(x^4-b)$, it means $2$ is a root of $x^4-b$, whence $b=2^4$.

Indeed, you easily check that $$x^5-16x=(x^2-2x)(x^3+2x^2+4x+8).$$

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Proof $1\!:\ $ $x(x\!-\!2)\mid xf \overset{{\rm cancel}\ x}\iff x\!-\!2\mid f \!\!\!\overset{\rm\large\color{#c00}{RT}\!\!\!}\iff f(2)=0 \overset{f = x^4-b\!}\iff b=2^4\,$ by $\color{#c00}{\small \rm RT}$ = Rem Theorem.
Proof $2\!:\ 0 = xf \bmod x(x\!-\!2) \overset{\color{#90f}{\rm DL}}=\, \color{#0a0}x(f \bmod x\!-\!2) \overset{\rm\large\color{#c00}{RT^{\phantom{|}}}_{\phantom |}\!\!\!}= x f(2)\! \iff\! f(2)=0\!\iff \ldots$
upon applying $\color{#90f}{\rm DL}\!: \ gf\bmod gh^{\phantom{|^{|^|}}}\!\!\! =\, \color{#0a0}g(f\bmod h)\,$ to factor out $\,\color{#0a0}{g\!=\!x},\,$ $\color{#90f}{\rm DL}$ = mod Distrib. Law.

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