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Thread: Irreduclible polynomial

  1. #1
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    Irreduclible polynomial

    Can somebody please check if the follwoing argument is correct:

    Show that x^4-4x-1 is irrdeucible over rationals.

    By the rational root theorem it has no rational roots.

    Assume that
    x^4-4x-1 = (x^2+ax+b)(x^2+cx+d) then

    a+c=0, d+ac+b=0 ad+bc=-4 bd=-1.


    a=-c implies that d+b-c^2=0
    c(b-d)=-4 bd=-1

    c=-4 or b-d=-4.
    Because bd=-1 it must hold that b=1,d=-1 or b=-1, d=1.

    c=-4 shows that d+b=16 which is a contradiction.
    if b-d=-4 we also get a contradiction. So the polynomial is irreducible.

    Is it correct?
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  2. #2
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    Quote Originally Posted by peteryellow View Post
    a=-c implies that d+b-c^2=0
    c(b-d)=-4 bd=-1

    c=-4 or b-d=-4.
    Because bd=-1 it must hold that b=1,d=-1 or b=-1, d=1.
    Since bd=-1 it means b=1,d=-1 or b=-1,d=1. In the first case it means b-d = 2 and so from c(b-d)=-4\implies c=-2. In the second case it means b-d =-2 and so from c(b-d)=-4\implies c=2. But we must have c^2 = b+d however, b+d=0 and c^2=4 in both cases. Thus, it is impossible to find such integers.
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  3. #3
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    Quote Originally Posted by peteryellow View Post
    Can somebody please check if the follwoing argument is correct:

    Show that x^4-4x-1 is irrdeucible over rationals.

    By the rational root theorem it has no rational roots.

    Assume that
    x^4-4x-1 = (x^2+ax+b)(x^2+cx+d) then

    a+c=0,\ d+ac+b=0\ ad+bc=-4\ bd=-1.


    a=-c implies that d+b-c^2=0,
    c(b-d)=-4, bd=-1 Correct as far as here!

    c=-4 or b-d=-4. NO!!!
    Because bd=-1 it must hold that b=1,d=-1 or b=-1, d=1.

    c=-4 shows that d+b=16 which is a contradiction.
    if b-d=-4 we also get a contradiction. So the polynomial is irreducible.

    Is it correct?
    My solution: from b+d=c^2 and bd=-1, deduce that b and d are the solutions of the quadratic equation t^2 - c^2t -1=0. But these solutions are t = \tfrac12(c^2\pm\sqrt{c^2+4}). It follows that b-d = \pm\sqrt{c^2+4}, and so c\sqrt{c^2+4} = \pm4. Therefore c^2(c^2+4) = 16. This is a quadratic equation for c^2, with irrational solutions c^2 = -2\pm\sqrt{20}. So there is no rational solution for c.
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