1. ## Polynomials

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

2. Hello NCA !
Spoiler:

$\displaystyle x \times q(x) \times p(x+2)=(x-1) \times p(x) \times q(x+2)$

This is equivalent to :

$\displaystyle \frac{x}{x - 1} = \frac{p(x) \times q(x + 2)}{q(x) \times p(x + 2)}$

By expanding :

$\displaystyle x^2 - x = \frac{p(x) \times q(x + 2)}{q(x) \times p(x + 2)}$

The product of two polynomials is a polynomial. However, the division of two polynomials is never a polynomial except when the polynomial in the denominator has degree zero (which is assumed not being the case). Since $\displaystyle x^2 - x$ is a polynomial, there exists no polynomials $\displaystyle p(x)$, $\displaystyle q(x)$ that satisfy the statement.

Is this enough as a proof, or is it invalid ?

3. 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 .

4. Originally Posted by Bacterius
Hello NCA !
Spoiler:

$\displaystyle x \times q(x) \times p(x+2)=(x-1) \times p(x) \times q(x+2)$

This is equivalent to :

$\displaystyle \frac{x}{x - 1} = \frac{p(x) \times q(x + 2)}{q(x) \times p(x + 2)}$

By expanding :

$\displaystyle x^2 - x = \frac{p(x) \times q(x + 2)}{q(x) \times p(x + 2)}$

The product of two polynomials is a polynomial. However, the division of two polynomials is never a polynomial except when the polynomial in the denominator has degree zero (which is assumed not being the case). Since $\displaystyle x^2 - x$ is a polynomial, there exists no polynomials $\displaystyle p(x)$, $\displaystyle q(x)$ that satisfy the statement.

Is this enough as a proof, or is it invalid ?
$\displaystyle \frac{x^2-1}{x-1}=x+1$