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Thread: Polynomials

  1. #1
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    Polynomials

    [$\displaystyle \star$] True or False: There exist non-zero polynomials $\displaystyle p(x), q(x) \in \mathbb{C}[x]$ such that $\displaystyle xq(x)p(x+2)=(x-1)p(x)q(x+2).$
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  2. #2
    Super Member Bacterius's Avatar
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    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 ?
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  3. #3
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    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 .
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bacterius View Post
    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 ?
    $\displaystyle \frac{x^2-1}{x-1}=x+1$
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