<s>[INST] Find all polynomials with real coefficients that satisfy $(x^2-6x+8)P(x)=(x^2+2x)P(x-2)$ 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?
 [/INST] 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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