[$\displaystyle \star$] True or False: There exist non-zero polynomials $\displaystyle p(x), q(x) \in \mathbb{C}[x]$ such that $\displaystyle xq(x)p(x+2)=(x-1)p(x)q(x+2).$
Given that $\displaystyle xq(x)p(x+2) = (x-1)p(x)q(x+2) ~ ... ~ (1)$
By replacing $\displaystyle x $ with $\displaystyle x-2 $ ,
we obtain :
$\displaystyle (x-2)q(x-2)p(x) = (x-3)p(x-2)q(x) $
then multiply it by $\displaystyle (1) $
$\displaystyle xq(x)p(x+2)\cdot(x-2)q(x-2)p(x) = (x-1)p(x)q(x+2) \cdot (x-3)p(x-2)q(x) $
There are $\displaystyle p(x) $ and $\displaystyle q(x) $ on the both sides , delete them . The remaining is :
$\displaystyle x(x-2)p(x+2)q(x-2) = (x-1)(x-3)p(x-2)q(x+2) ~...~(2) $
Repeat this procedure again ,
we obtain
$\displaystyle (x-2)(x-4)p(x)q(x-4) = (x-3)(x-5)p(x-4)q(x) $
then mutiply it by $\displaystyle (1) $ , the reference equation .
$\displaystyle x(x-2)(x-4)p(x+2)q(x-4) = (x-1)(x-3)(x-5)p(x-4)q(x+2) ~...~(3) $
I find that comparing with the previous equation , eg $\displaystyle (2) $
there is an extra linear factor on each side in $\displaystyle (3) $ and a change $\displaystyle p(x-2)$ to $\displaystyle p(x-4) $ and $\displaystyle q(x-2) $ to $\displaystyle q(x-4) $ , so I assume it is also true that :
$\displaystyle x(x-2)(x-4)...(x-2n)p(x+2)q(x-2n) = (x-1) $$\displaystyle (x-3)(x-5)...(x-2n-1)p(x-2n)q(x+2) $
Then at least one of them $\displaystyle p(x) , q(x) $ contains many factors , so it is not a polynomial .