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Math Help - Solutions of an equation

  1. #1
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    Solutions of an equation

    I need some help with the following problem:

    (a) Prove that the equation x^2 + 23y^2 = 41 does not have solutions in integers.

    (b) If (1/3, 4/3) and (9/4, 5/4) are two rational solutions of this equation, find a solution to this equation in \mathbb{Z}_{568} and another one in \mathbb{Z}_{657}.

    I have not encountered this type of problems before, so I'm unsure how to get started on either parts. Any help to get me started on them is greatly appreciated.
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  2. #2
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    If youl let \displaystyle y = 0, then you have

    \displaystyle x^2 = 41.

    There is not an integer solution to this equation.


    If you let \displaystyle y = \pm 1, then you have

    \displaystyle x^2 + 23 = 41

    \displaystyle x^2 = 18.

    There is not an integer solution to this equation.


    If you let \displaystyle y = \pm 2, you have

    \displaystyle x^2 + 92 = 41

    \displaystyle x^2 = -51 which also does not have any integer (or real) solutions.

    It should also be clear that any other integer will also give \displaystyle x^2 = \textrm{something negative}, which will not give any integer (or real) solution.
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  3. #3
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    Quote Originally Posted by demode View Post
    I need some help with the following problem:

    (a) Prove that the equation x^2 + 23y^2 = 41 does not have solutions in integers.

    (b) If (1/3, 4/3) and (9/4, 5/4) are two rational solutions of this equation, find a solution to this equation in \mathbb{Z}_{568} and another one in \mathbb{Z}_{657}.

    I have not encountered this type of problems before, so I'm unsure how to get started on either parts. Any help to get me started on them is greatly appreciated.
    For part (a), y^2 can only be 0 or 1. (Otherwise x will not be real.)

    So, we get two cases:

    1) x^2=41

    2) x^2+23=41 \implies  x^2=18

    Clearly, neither of these equations have integral solutions.

    I have no idea about part (b).
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  4. #4
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    For part (b) , observe that

    568 = 2^3 \cdot 71

    and

    657 = 3^2 \cdot 73

    Therefore, gcd(3,568) = 1 and gcd(4,657) = 1. It follows that each of the equations

    3x \equiv 1 (mod \ 568)

    and

    4y \equiv 1 (mod \ 657)

    has a solution in \mathbb{Z}. Can you take it from there?
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  5. #5
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    Quote Originally Posted by Petek View Post
    For part (b) , observe that

    568 = 2^3 \cdot 71

    and

    657 = 3^2 \cdot 73

    Therefore, gcd(3,568) = 1 and gcd(4,657) = 1. It follows that each of the equations

    3x \equiv 1 (mod \ 568)

    and

    4y \equiv 1 (mod \ 657)

    has a solution in \mathbb{Z}. Can you take it from there?
    How do you solve this kind of equations (like 4y \equiv 1 (mod \ 657) for example) without checking all different elements in \mathbb{Z}_{657}? Do you somehow use the prime power factorization?
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  6. #6
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    Quote Originally Posted by demode View Post
    How do you solve this kind of equations (like 4y \equiv 1 (mod \ 657) for example) without checking all different elements in \mathbb{Z}_{657}? Do you somehow use the prime power factorization?
    If you're studying a number theory textbook (or have access to one), it should have a section on solving linear congruences of the form

    ax \equiv b (mod \ m)

    Solving the above congruence is equivalent to solving the diophantine equation

    ax - my = b

    which can be solved, for example by the Euclidean algorithm or continued fractions. Another way to solve the equation

    4y \equiv 1 (mod \ 657)

    is to observe that

    \phi(657) = \phi(3^2) \cdot \phi(73) = 3 \cdot 72 = 216

    Use Euler's theorem to conclude that

    4^{\phi(657)} \equiv 1 (mod \ 657)

    and so

    4 \cdot 4^{215} \equiv 1 (mod \ 657)

    Then simplify 4^{215} mod 657.
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