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Math Help - No integer Solutions

  1. #1
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    Unhappy No integer Solutions

    Prove that 15x^2 -7y^2=9 has no integer solutions.

    I have understood a few steps

    15x^2 -7y^2=9 => y=3y1 => 5x^2-21y^2=3 => x-3x1 => 15x1^2-7y1^2=1 =>y1^2=2 mod 3

    But this contradiction how???
    If there is any other way plz tell.

    Thanx
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: No integer Solutions

    Quote Originally Posted by Learner248 View Post
    Prove that 15x^2 -7y^2=9 has no integer solutions.

    I have understood a few steps

    15x^2 -7y^2=9 => y=3y1 => 5x^2-21y^2=3 => x-3x1 => 15x1^2-7y1^2=1 =>y1^2=2 mod 3

    But this contradiction how???
    If there is any other way plz tell.

    Thanx
    I'm sorry that's a little hard to make out, presumably what it says is that if x,y were solutions then 7y^2=9-15x^2\equiv 0\text{ mod }3 and so 3\mid \y^2 but since 3 is prime and 3\nmid 7 we may conclude that 3\mid y and so y=3y' and so we must have that 15x^2-63y'^2=9 but in the same spirit we see that this implies 9\mid 15x^2 and so 3\mid 5x^2 and since 3 is prime and 3\nmid we may conclude that 3\mid x and so x=3x' and so our equation becomes, after dividing both sides by 9, the following 15x'^2-7y'^2=1 but this says that y'^2\equiv 7y'^2\equiv -1\equiv 2\text{ mod }3 but it's evident that 2 isn't a quadratic residue \text{ mod }3/
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  3. #3
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    Re: No integer Solutions

    Hello, Learner248!

    I am puzzled, too.


    Prove that: 15x^2 -7y^2\:=\:9 has no integer solutions.

    I have understood a few steps

    15x^2 -7y^2\:=\:9

    y=3y_1 \quad\Rightarrow\quad 5x^2-21y^2\:=\:3

    x-3x_1\quad\Rightarrow\quad 15x_1^2-7y_1^2\:=\:1

    y_1^2\:=\:2\text{ (mod 3)}

    But this contradiction how?

    If there is any other way, plz tell.

    We have: . 15x^2 - 7y^2 \:=\:9

    Since 15x^2 and 9 are multiples of 3,
    . . then 7y^2 must be a multiple of 3.
    Hence, y is a multiple of 3: . y \:=\:3y_1

    The equation becomes: . 15x^2 - 7(3y_1)^2 \:=\:9 \quad\Rightarrow\quad 15x^2 - 63y_1^2 \:=\:9
    . . Divide by 3: . 5x^2 - 21y_1^2 \:=\:3

    Since 21y_1^2 and 3 are multiples of 3,
    . . then 5x^2 must be a multiple of 3.
    Hence, x is a multiple of 3: . x \:=\:3x_1

    The equation becomes: . 5(3x_1)^2 - 21y_1^2 \:=\:3 \quad\Rightarrow\quad 45x_1^2 - 21y_1^2 \:=\:3
    . . Divide by 3: . 15x_1^2 - 7y_1^2 \:=\:1

    And I have no idea where they got: y_1^2 = 2\text{ (mod 3)}
    . . It doesn't seem to be true.


    I think I can finish the proof . . .

    We have: . 15x_1^2 - 7y_1^2 \:=\:1

    Then: . y_1^2 \:=\:\frac{15x_1^2-1}{7} \quad\Rightarrow\quad y_1^2 \:=\:2x_1^2 + \frac{x_1^2-1}{7} .[1]

    Since y_1^2 is an integer, then x_1^2-1 must be a multiple of 7.
    . . That is: . x_1^2-1 \:=\:7k \quad\Rightarrow\quad x_1^2 \:=\:7k+1

    Substitute into [1]: . y_1^2 \:=\:2(7k+1) + \frac{(7k+1)-1}{7}
    . . which simplfies to: . y_1^2 \:=\:15k+2


    For integer values of k,\:y_1^2 will end in 2 or 7.

    But squares must end in \{0,1,4,5,6,9\}

    There is our contradiction!

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  4. #4
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    Re: No integer Solutions

    Quote Originally Posted by Learner248 View Post
    Prove that 15x^2 -7y^2=9 has no integer solutions.

    I have understood a few steps

    15x^2 -7y^2=9 \;\Rightarrow\; \ldots \;\Rightarrow\; y_1^2=2 \pmod 3

    But this contradiction how???
    There are no numbers with square congruent to 2 mod 3. In fact, every integer is congruent to 0, 1 or 2 (mod 3), and

    0^2\equiv0\pmod3,\quad 1^2\equiv1\pmod3,\quad 2^2\equiv1\pmod3.

    So the equation y_1^2=2 \pmod 3 has no solutions.
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  5. #5
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    Re: No integer Solutions

    Thanks Everybody

    I figured it out a few minutes later after I posted it.

    Anyways

    Thanx a lot
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