Polynomials

**NonCommAlg** · March 1st 2010, 11:15 PM

[ $\star$ ] True or False: There exist non-zero polynomials $p(x), q(x) \in \mathbb{C}[x]$ such that $xq(x)p(x+2)=(x-1)p(x)q(x+2).$

**Bacterius** · March 2nd 2010, 08:38 PM

Hello NCA !

Spoiler:

**simplependulum** · March 2nd 2010, 10:54 PM

Given that $xq(x)p(x+2) = (x-1)p(x)q(x+2) ~ ... ~ (1)$

By replacing $x$ with $x-2$ ,
we obtain :

$(x-2)q(x-2)p(x) = (x-3)p(x-2)q(x)$

then multiply it by $(1)$

$xq(x)p(x+2)\cdot(x-2)q(x-2)p(x) = (x-1)p(x)q(x+2) \cdot (x-3)p(x-2)q(x)$

There are $p(x)$ and $q(x)$ on the both sides , delete them . The remaining is :

$x(x-2)p(x+2)q(x-2) = (x-1)(x-3)p(x-2)q(x+2) ~...~(2)$

Repeat this procedure again ,

we obtain

$(x-2)(x-4)p(x)q(x-4) = (x-3)(x-5)p(x-4)q(x)$

then mutiply it by $(1)$ , the reference equation .

$x(x-2)(x-4)p(x+2)q(x-4) = (x-1)(x-3)(x-5)p(x-4)q(x+2) ~...~(3)$

I find that comparing with the previous equation , eg $(2)$
there is an extra linear factor on each side in $(3)$ and a change $p(x-2)$ to $p(x-4)$ and $q(x-2)$ to $q(x-4)$ , so I assume it is also true that :

$x(x-2)(x-4)...(x-2n)p(x+2)q(x-2n) = (x-1)$ $(x-3)(x-5)...(x-2n-1)p(x-2n)q(x+2)$

Then at least one of them $p(x) , q(x)$ contains many factors , so it is not a polynomial .