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Equation about polynomials that implies polynomials are zero

  1. Dec 25, 2008 #1
    Hi everyone,

    I have to demonsrate that for every real polynomial, P Q and R, I have :

    P²=X(Q²+R²) ==imply==> P=Q=R=0

    Using degrees, we can easily demonsrate the above. However, I'm looking for another way, without using that.
     
  2. jcsd
  3. Dec 25, 2008 #2

    Tom Mattson

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    Re: polynomials

    What is X? :confused:
     
  4. Dec 25, 2008 #3
    Re: polynomials

    X is the variable ! We can rewrite the problem this way :
    if P(X), Q(X) and R(X) belongs to R[X], then writing P²(X) = X(Q²(X) + R²(X)) means P(X)=Q(X)=R(X)=0

    Someone adviced me to start with demonstrating that constant coefficients are equal to 0 ? :s
     
  5. Dec 25, 2008 #4

    HallsofIvy

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    Re: polynomials

    The problem still makes no sense. Surely you don't mean "for every real polynomial, P Q and R" and then say "P= Q= R= 0".

    Suppose Q= R= x and P= 2x3. Does that contradict what you are trying to prove? Are you assuming that P, Q, and R are have the same degree?
     
  6. Dec 25, 2008 #5
    Re: polynomials

    My bad ! Here's the exact question :
    "Demonstrate that if P, Q, and R belong to R[X], therefore P² - XQ² = XR² imply that P=Q=R=0"

    Big logic mistake in my first post -.- sorry
     
  7. Dec 26, 2008 #6

    HallsofIvy

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    Re: polynomials

    I think the "degree argument" you mentioned is still the simplest and best way to go. If n is the higher degree of Q and R, then the degree of xQ2+ R2 is 2n+ 1 so the degree of P would have to be (2n+1)/2, not an integer.
     
  8. Dec 26, 2008 #7

    Hurkyl

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    Re: polynomials

    At the risk of pointing out the obvious, [itex]f(x)^2 \geq 0[/itex] for every real x....
     
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