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Thread: Number Theory Help

  1. #1
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    Number Theory Help

    I need help solving a problem for my discrete class.

    The problem is : Prove no integer solutions of x and y satisfy the equation  3x^2 + 5y^2 = 19 . Hint: Consider this equation in modulo 5.

    My attempt:
    Assume  3x^2 + 5y^2 = 19
    Thus, in mod 5,

     3x^2 + 5y^2 \equiv 19

     3x^2 + 5y^2 \equiv 4
    Note that 5y^2 in mod 5 will always be 0, since it will be a multiple of 5, thus:


     3x^2 \equiv 4

    Now I broke it case by case to see what x^2 will have to be to satisfy the above requirement.
    I found that

     x^2 \equiv 3 in order for
     3x^2 \equiv 4

     x^2 \equiv 3 implies  (\exists k \in Z )(5k+3 = x^2)

    The following steps may be useless and/or incorrectly derived, but i dont know what else to do:

    if 5k+3 = x^2 , then  5k = x^2 - 3

    What now?
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  2. #2
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    Re: Number Theory Help

    If x \equiv 0 \pmod{5} then x^2 \equiv  0 \pmod{5}.
    If x \equiv 1 \pmod{5} then x^2 \equiv  1 \pmod{5}.
    If x \equiv 2 \pmod{5} then x^2 \equiv  4 \pmod{5}.
    If x \equiv 3 \pmod{5} then x^2 \equiv  4 \pmod{5}.
    If x \equiv 4 \pmod{5} then x^2 \equiv  1 \pmod{5}.
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  3. #3
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    Re: Number Theory Help

    there's probably a more formal method but you can notice that

    $x^2 \pmod{5} = (x \pmod{5})(x \pmod{5}) \pmod{5}$

    so we only have to look at $x \in \{0,1,2,3,4\}$

    $x^2 \pmod{5} \in \{0,1,4\} \Rightarrow 3x^2 \pmod{5} \in \{0,2,3\}$

    $4 \not \in \{0,2,3\}$

    so the equation has no solutions mod 5

    if there was a general solution it would be a solution mod any integer
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