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Math Help - Binary Quadratic Forms

  1. #1
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    Question Binary Quadratic Forms

    a) [solved] Show that x^2 + 5y^2 and 2x^2 + 2xy + 3y^2 are the only reduced positive definite binary quadratic forms of discriminant -20. Show that the first of these forms does not represent 2, but that the second one does, so that these forms are inequivalent, and hence that the class number H(-20)=2. Show that a prime p is represented by at least one of these forms if and only if p=2, p=5, or p≡1,3,7, or 9 (mod20). [solved]

    b) [solved] Show that if a number n is represented by a binary quadratic form f(x,y) = ax^2 + bxy + cy^2 of discriminant d, then 4an is a square modulo |d|. [solved]

    c) Using part b, show that if a prime p is represented by the form x^2 + 5y^2, then the Legendre symbol (p/5) =1 or 0, and that if p is a prime represented by the form 2x^2 + 2xy + 3y^2, then (p/5)= -1.

    d) By combining parts a & c, show that an odd prime p is represented by the form x^2 + 5y^2 if and only if p=5 or p≡1 or 9 (mod20), and that a prime p is represented by the form 2x^2 + 2xy + 3y^2 if and only if p=2 or p≡3 or 7 (mod 20).
    ===================================

    I proved parts a & b, but I'm stuck on parts c & d.
    a) An odd prime p is represented by at least one of these forms of discriminant -20
    <=> p|(-20) or (-20/p)=1
    <=> p=5 or (-20/p)=1
    <=> p=5 or (-1)^{(p - 1)/2}(p/5)=1
    Now (-1)^{(p - 1)/2}(p/5)=1 <=> (-1)^{(p - 1)/2}=(p/5)=1 or (-1)^{(p - 1)/2}=(p/5)= -1
    (-1)^{(p - 1)/2}=(p/5)=1 <=> p≡1(mod 4) and p≡1,4(mod 5) <=>p≡1,9(mod 20)
    (-1)^{(p - 1)/2}=(p/5)= -1 <=> p≡3(mod 4) and p≡2,3(mod 5) <=>p≡3,7(mod 20)

    c) If p is a prime represented by x^2 + 5y^2, then by part b, 4p≡ z^2 (mod 20) has a solution.
    z must be even, so let z=2k
    => 4p≡ 4k^2 (mod 20) => p≡ k^2 (mod 5)
    => 5|p or (p/5)=1 => p=5 or (p/5)=1 => (p/5)=0 or 1.

    If p is a prime represented by 2x^2 + 2xy + 3y^2, then by part b, 8p≡ z^2 (mod 20) has a solution.
    z must be even, so let z=2k
    => 8p≡ 4k^2 (mod 20) => 2p≡ k^2 (mod 5)
    => 5|2p or (2p/5)=1 => 5|p or (2p/5)=1
    => p=5 or (2p/5)=1
    => p=5 or (p/5)= -1
    => (p/5)= 0 or -1 which is NOT what we want, but I can't find where my mistake is...

    d) I'm not sure how to "combine" the information in parts a & c. In particular, we are required to prove "if and only if" in part d. How can we prove it?

    Can someone help me, please?
    Thank you!


    [also under discussion in math links forum]
    Last edited by kingwinner; March 21st 2010 at 12:54 PM.
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  2. #2
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    So for part c, I think we have to show that the case p=5 is impossible.

    In short, my question just translates to the following:
    How can we prove that  2x^{2}+2xy+3y^{2} =5 has no integer solutions?

    Thanks!
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  3. #3
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    Quote Originally Posted by kingwinner View Post
    So for part c, I think we have to show that the case p=5 is impossible.

    In short, my question just translates to the following:
    How can we prove that  2x^{2}+2xy+3y^{2} =5 has no integer solutions?

    Thanks!
    If we take the polynomial in terms of x, the discriminant of 2x^2 + 2xy + 3y^2 - 5 = 0 is 20(2 - y^2). In order for x to be a real number, this value must be positive, and in order for x to be an integer, this value must be a square. The only integer values for y such that x is real are -1, 0, and 1, and none of these values makes the discriminant a square. Hence 2x^2 + 2xy + 3y^2 = 5 has no integer solutions.
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by kingwinner View Post
    So for part c, I think we have to show that the case p=5 is impossible.

    In short, my question just translates to the following:
    How can we prove that  2x^{2}+2xy+3y^{2} =5 has no integer solutions?

    Thanks!
    Consider the equation modulo  4

     5 \equiv 1 \mod{4}

    There are  16 pairings of  (x,y) consider for  f(x,y)=2x^2+2xy+3y^2 modulo  4 .

    It turns out they all come out to be either  0,2, or 3 .

    So since we cant find  (x,y) such that  f(x,y) \equiv 1 \mod{4} we are done.
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  5. #5
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    Quote Originally Posted by chiph588@ View Post
    Consider the equation modulo  4

     5 \equiv 1 \mod{4}

    There are  16 pairings of  (x,y) consider for  f(x,y)=2x^2+2xy+3y^2 modulo  4 .

    It turns out they all come out to be either  0,2, or 3 .

    So since we cant find  (x,y) such that  f(x,y) \equiv 1 \mod{4} we are done.
    Why did you choose mod 4 (rather than mod something else)? How did you get this inspiration to work with mod 4? It's kind of mysterious to me, and I have no idea how to come up with inspirations like this...Can you explain the thinking behind it?


    Also, the logic of your proof is the following, right?
    The conrguence 2x^2 + 2xy +3y^2 ≡ 5 (mod 4) has no solutions, thus the equation 2x^2 + 2xy +3y^2 = 5 has no integer solutions, right?


    Thank you!
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by kingwinner View Post
    Why did you choose mod 4 (rather than mod something else)? How did you get this inspiration to work with mod 4? It's kind of mysterious to me, and I have no idea how to come up with inspirations like this...Can you explain the thinking behind it?


    Also, the logic of your proof is the following, right?
    The conrguence 2x^2 + 2xy +3y^2 ≡ 5 (mod 4) has no solutions, thus the equation 2x^2 + 2xy +3y^2 = 5 has no integer solutions, right?


    Thank you!
    Well I started with mod 3 and when that didn't work, I just moved up the ladder. Luckily mod 4 worked.

    And your reasoning is correct too.
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  7. #7
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    Smile

    I see.

    Is there any alternative way to prove that 2x^{2}+2xy+3y^{2}=5 has no integer solutions?

    Say, by completing the square?
     <br />
(x+y)^{2}+x^{2}+2y^{2} = 5<br />
    Is it possible to eliminate some cases from here and show the remaining cases do not give 5?
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by kingwinner View Post
    I see.

    Is there any alternative way to prove that 2x^{2}+2xy+3y^{2}=5 has no integer solutions?

    Say, by completing the square?
     <br />
(x+y)^{2}+x^{2}+2y^{2} = 5<br />
    Is it possible to eliminate some cases from here and show the remaining cases do not give 5?
    Since all three terms are positive,  |y|\leq 1 and  |x|\leq 2 .

    You could probably find more constrains by looking at  x+y , but the above two constraints gives us 15 cases to check, which is good enough for me .
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