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Math Help - Quadratic Number Field

  1. #1
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    Quadratic Number Field

    Let D \in \mathbb{Z} be square free. Consider K = \{  x| ax^2+bx+c = 0, a,b,c \in \mathbb{Z} \}.

    For which conditions on a,b,c\in \mathbb{Z} is K=  \{ u + v\sqrt{D}: u,v \in \mathbb{Q} \}?
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  2. #2
    Senior Member Dinkydoe's Avatar
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    With the abc-formula we have K= \left\{\frac{-b}{2a}\pm \frac{1}{2a}\sqrt{b^2-4ac}\right\}

    Hence we must have kD=b^2-4ac for some square k
    ( \sqrt{k}\in \mathbb{Q})
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  3. #3
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    Could you enlarge on that, what are the conditions on a,b,c?
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  4. #4
    Senior Member Dinkydoe's Avatar
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    Ok.

    So we can choose k=p^2\in \mathbb{Z} free.

    Then we choose b\in \mathbb{Z} such that ac=\frac{b^2-kD}{4}\in \mathbb{Z} is not prime.

    Then a,c divide this number.
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  5. #5
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    So you say that we choose a and c arbitrarily, such that ac is not a prime, and then b needs to be chosen such that 4ac = b^2 -kD, for some k=p^2 (p \in \mathbb{Z})

    Is it that what you mean? And why is this what we want??
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  6. #6
    Senior Member Dinkydoe's Avatar
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    with the abc-formula we have that K consists of the numbers \frac{-b}{2a}\pm \frac{1}{2a}\sqrt{b^2-4ac}

    To make sure K has the desired property we must have \sqrt{b^2-4ac}=p\sqrt{D} for some number p\in \mathbb{Z}.

    That is, b^2-4ac= p^2D.

    This number p^2\in \mathbb{Z} can ofcourse be freely chosen.

    Then we must choose a,b,c\in\mathbb{Z} that satisfy this relation. Thus we must choose a b\in \mathbb{Z} such that

    there exists a composite number ac\in\mathbb{Z} that satisfies b^2 = p^2D+4ac

    wich is, the same as saying \frac{b^2-p^2D}{4} must be a composite number. Only then we can choose a,c\in \mathbb{Z} such that ac=\frac{b^2-p^2D}{4}.

    And we've found  a,b,c such that K has the desired property.
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  7. #7
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    Why does ac need to be composite?
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  8. #8
    Senior Member Dinkydoe's Avatar
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    Why does ac need to be composite?
    How else can we find a,c\in \mathbb{Z}?

    If ac=p prime then a,c can not be integers.
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  9. #9
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Dinkydoe View Post
    How else can we find a,c\in \mathbb{Z}?

    If ac=p prime then a,c can not be integers.
     a=1,\; c=p
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  10. #10
    Senior Member Dinkydoe's Avatar
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    Ah shoot, big blunder here. Shameful moment here xp

    Thanks chiph

    Forget everything about ac being composite. I'm sorry for this error ;p


    The rest you can take for granted.
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  11. #11
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    So we just need to choose a,b,c such that \sqrt{b^2-4ac}=p\sqrt{D} for some p \in \mathbb{Z} and thats all?
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  12. #12
    Senior Member Dinkydoe's Avatar
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    no, I Edited my post ;p

    with the abc-formula we have that consists of the numbers

    To make sure has the desired property we must have for some number .

    That is, .

    This number can ofcourse be freely chosen.

    Then we must choose that satisfy this relation. Thus we must choose a such that

    there exists a number that satisfies

    wich is, the same as saying \in \mathbb{Z}. Only Then we can choose such that .

    And we've found such that has the desired property.
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  13. #13
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    I don't see the difference. Which condition do you have except b^2-4ac=p^2 D for  p \in \mathbb{Z}<br />
.

    I mean once this holds for integers a,b,c, ac \in \mathbb{Z} trivially.
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  14. #14
    Senior Member Dinkydoe's Avatar
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    No, it's not different. It's the same, but you wanted more details right ;p?
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