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Math Help - is this a 'safe' assumption?

  1. #1
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    is this a 'safe' assumption?

    Hi,

    If I have the following equation

    8x^2-4y^2=z

    and it is known that x and y are both non-equal, non-zero, even integers

    then z will even and can be written z=2u

    so that 8x^2-4y^2=2u

    It is also required that z be a square integer

    so it can be rewritten z=2u=4v^2

    So with the above constraints I believe that

    8x^2-4y^2=z

    can be rewritten

    8x^2-4y^2=4v^2

    so

    2x^2-y^2=v^2

    Have I made any fundamental mistakes anywhere?

    Many thanks

    Pro
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  2. #2
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    Re: is this a 'safe' assumption?

    Given all the assumptions you made, that is correct.
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  3. #3
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    Re: is this a 'safe' assumption?

    Thanks HallsofIvy, glad to have someone check it over

    Pro
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  4. #4
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    Re: is this a 'safe' assumption?

    What if y ≥ 2x ?
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  5. #5
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    Re: is this a 'safe' assumption?

    Quote Originally Posted by SammyS View Post
    What if y ≥ 2x ?
    Thanks SammyS.

    That's true but the fact that z is a square implies  x>y/2
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  6. #6
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    Re: is this a 'safe' assumption?

    Quote Originally Posted by procyon View Post
    8x^2-4y^2=z
    and it is known that x and y are both non-equal, non-zero, even integers
    then z will even and can be written z=2u
    so that 8x^2-4y^2=2u
    x and y can also be odd ... multiplications by evens (4,8 in your case)
    always give even results....
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  7. #7
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    Re: is this a 'safe' assumption?

    Quote Originally Posted by procyon View Post
    Thanks SammyS.

    That's true but the fact that z is a square implies  x>y/2
    Usually the requirements are written at the beginning of the problem, so I missed the fact that z can be written as a square.
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  8. #8
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    Re: is this a 'safe' assumption?

    Quote Originally Posted by Wilmer View Post
    x and y can also be odd ... multiplications by evens (4,8 in your case)
    always give even results....
    Thanks Wilmer.

    That's true but x and y odd doesn't fit with other, simultaneous, equations I'm working with, so I thought I'd just state they're even in this case. x and y odd is valid though for a seperate set of equations.

    Pro
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  9. #9
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    Re: is this a 'safe' assumption?

    Quote Originally Posted by SammyS View Post
    Usually the requirements are written at the beginning of the problem, so I missed the fact that z can be written as a square.
    Hi SammyS,

    Yes, sorry about that. I was thinking whilst wrting instead of before

    Pro
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  10. #10
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    Re: is this a 'safe' assumption?

    Quote Originally Posted by procyon View Post
    That's true but x and y odd doesn't fit with other, simultaneous, equations I'm working with, so I thought
    I'd just state they're even in this case. x and y odd is valid though for a seperate set of equations.
    Disagree; x and y odd will work with the equations you've shown.
    No separate set is required...let's see if Sammy agrees...
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  11. #11
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    Re: is this a 'safe' assumption?

    Quote Originally Posted by Wilmer View Post
    Disagree; x and y odd will work with the equations you've shown.
    No separate set is required...let's see if Sammy agrees...
    I think if you re-read my previous reply, you'll find that I did actually agree with you on this

    I'm actually saying that this equation was derived by setting other values to even

    i.e.

    a=2b guarantees a is a positive even integer if b is any integer > 0
    a=2b+1 guarantees a is positive odd integer if b is any integer ≥ 0

    You might see the whole thing soon. At the minute I seem to have

    2(odd integer) - (different [or same] odd integer) = even integer

    which would prove what I'm trying to calculate is impossible.

    [edit]
    damn, turned out to be odd, lol
    [/edit]

    Pro
    Last edited by procyon; August 19th 2011 at 05:04 AM.
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