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Find all polynomials with real coefficients that satisfy $$(x^2-6x+8)P(x)=(x^2+2x)P(x-2)\forall x\in\Bbb R$$

My work;

$$\frac{P(x)}{P(x-2)}=-\frac{4}{x-2}+\frac{12}{x-4}+1\tag{1}$$

$$\frac{P(x-2)}{P(x)}=-\frac{12}{x+2}+\frac{4}{x}+1\tag{2}$$

I also factorised the two known polynomial which didn't give anything useful. What should I extract from these two ratios?

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  • $\begingroup$ Since $0$ and $-2$ are roots of $x^2+2x$ and aren't roots of $x^2-6x+8$ we have that (using the LHS) $P(-2)=P(0)=0$, by other hand $4$ and $2$ are roots of $x^2-6x+8$ and aren't roots of $x^2+2x$, thus (using the RHS) $P(2)=P(4)=0$ $\endgroup$ – DiegoMath Nov 15 '14 at 17:40
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$$(x^2-6x+8)P(x)=(x^2+2x)P(x-2)\iff (x-2)(x-4)P(x)=(x+2)xP(x-2)$$ When $x=2,4$ LHS=0, so $RHS=0$, hence $x=0,2$ is a root for $P(x)$.

When $x=0,-2$ RHS=0, so $LHS=0$, hence $x=0,-2$ is a root for $P(x)$.

Write $P(x)=a(x)\cdot(x-2)x(x+2)$, plug in the assumption, we have $$a(x)\cdot(x-4)(x-2)^2x(x+2)=a(x-2)\cdot(x-4)(x-2)x^2(x+2)$$ $$\implies a(x)\cdot(x-2)=a(x-2)\cdot x$$ $$\implies a(x)=ax$$ For the last step, we first argue $a(x)$ can't have constant term, since $x=0$ is a root. Hence $a(x)=xb(x)$, which implies $b(x)=b(x−2)\forall x\in \mathbb{R}$.

Then we use the fact if polynomial $p(x)$ takes value $p_0$ infinitely many times (which means $p(x)-p_0=0$ has infinite zeros), then $p(x)$ is constant.

Hence $P(x)=ax^2(x-2)(x+2)$

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  • $\begingroup$ Why does $a(x)(x-2)=a(x-2)x$ imply $a(x)=ax$? Isn't $x-2$ the arguement of the function $a(\cdot)$? $\endgroup$ – UserX Nov 15 '14 at 17:52
  • $\begingroup$ @UserX First $a(x)$ can't have constant term, otherwise just set $x=0$ you will get a contradiction. Hence $a(x)=xb(x)$, if you divide $x(x-2)$ from both sides, you can get $b(x)=b(x-2)\forall x\in\mathbb{R}$, which implies $b(x)$ is constant. $\endgroup$ – John Nov 15 '14 at 18:06
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The relation is $(x-2)(x-4)P(x)=x(x+2)P(x-2)$ hence $$\frac{P(x)}{(x+2)x^2(x-2)}=\frac{P(x-2)}{x(x-2)^2(x-4)}.$$ Can you conclude?

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Let's exclude the trivial solution where $P$ is the zero polynomial.

Assume $z$ is a complex root of $P$ and $z\notin\{-2,0,2\}$. Then $z+2$ is a root of the right hand side, hence is either a root of $P$ (or $\in\{2,4\}$, but that is excluded). Also, $z$ is a root of th eleft hand side, hence $z-2$ is a root of $P$ (or $z\in\{0,-2\}$, which again is excluded). Hence for any complex root $z\notin2-2\mathbb N_0=\{\ldots,-4,-2,0,2\}$ we obtain infinitely many roots $z,z+2,z+4,\ldots$ of $P$; and for any complex root $z\notin-2+2\mathbb N_0=\{-2,0,2,4,\ldots\}$ we obtain infinitely many roots $z,z-2,z-4,\ldots$ of $P$. Both is absurd, hence all complex roots of $P$ are among$-2, 0, 2$. We conclude $P(x)=\alpha x^a(x-2)^b(x+2)^c$. Then (for $\alpha\ne0$) the original equation becomes $$(x-2)(x-4)\cdot x^a(x-2)^b(x+2)^c = x(x+2)\cdot (x-2)^a(x-4)^bx^c$$ i.e. $$(x-4)(x-2)^{b+1}x^a(x+2)^c = (x-4)^b(x-2)^ax^{c+1}(x+2)$$ hence $b=1, a=b+1, c+1=a, c=1$ and finally $$P(x)=\alpha x^2(x-2)(x+2).$$

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Hint: $$(x-4)(x-2)P(x)=x(x+2)P(x-2)$$ implies that $$P(x) = x(x+2)G(x).$$

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  • $\begingroup$ Sorry but I cannot continue. @upvoters Care to explain? $\endgroup$ – Did Nov 15 '14 at 17:48

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