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Math Help - Non-existence

  1. #1
    Super Member malaygoel's Avatar
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    Non-existence

    Show that there do not exist polynomials p(x) and q(x) of degree greater than or equal to 1 with integer co-efficients such that
    p(x)q(x)=x^5+ 2x+1.
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  2. #2
    Senior Member I-Think's Avatar
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    Re: Non-existence

    If p(x)q(x)=x^5+ 2x+1, then either
    Case 1
    p is of degree 1 and q is of degree 4,
    or
    Case 2
    p is of degree 2 and q is of degree 3

    I'll do case 1
    (ax+b)(cx^4+dx^3+ex^2+fx+g)=x^5+2x+1
    But the polynomials have integer coefficients, so a=c=\pm{1} and b=g=\pm{1}
    (I'm going to choose +1 for both to simplify things here)

    (x+1)(x^4+dx^3+ex^2+fx+1)=x^5+2x+1
    Expanding
    x^5+(d+1)x^4+(e+d)x^3+(e+f)x^2+(f+1)x+1=x^5+2x+1
    Comparing coefficients
    d+1=0, so d=-1
    e+d=0, so e=1
    e+f=0, so f=-1
    f+1=2, so f=1

    Contradiction has occurred, so Case 1 is impossible
    Case 2 is proven similarly
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  3. #3
    Super Member malaygoel's Avatar
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    Re: Non-existence

    I just wish to know if the following reasoning is correct?

    Working few steps as given by I-Think above:
    Quote Originally Posted by I-Think View Post
    If p(x)q(x)=x^5+ 2x+1, then either
    Case 1
    p is of degree 1 and q is of degree 4,
    or
    Case 2
    p is of degree 2 and q is of degree 3

    I'll do case 1
    (ax+b)(cx^4+dx^3+ex^2+fx+g)=x^5+2x+1
    But the polynomials have integer coefficients, so a=c=\pm{1} and b=g=\pm{1}
    (I'm going to choose +1 for both to simplify things here)

    (x+1)(x^4+dx^3+ex^2+fx+1)=x^5+2x+1
    Now, since x=-1 is a root of LHS but not of RHS, LHS cannot be equal to RHS for any values of d,e and f. Is it correct to conclude like this?
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  4. #4
    Senior Member I-Think's Avatar
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    Re: Non-existence

    Yes, it is a very correct and much shorter and neater way of answering the question, for case 1.
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Non-existence

    Maybe another solution(only highlights):

    x^5+ 2x+1=0 have only one real root (wolframalpha or calculus laws).

    Suppose that m is root of x^5+ 2x+1, now, we need to prove that m is not rational.

    Lets say that m=p/q , where p and q are integers.

    m^5+ 2m+1=(p/q)^5+ 2(p/q)+1=0

    ==>

    p^5+2p*q^4+q^5=0

    Integer solution only when p=q=0

    Contradiction! Therefore, m is irrational.
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