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Math Help - can this expression be a perfect square?

  1. #1
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    can this expression be a perfect square?

    Hi,

    Can anyone show that 4a^4+8a^3 b-12ab^3-7b^4 can be a perfect square if a and b are integers and a > b > 0 ?

    I think that a(a+2b) and b(12a+7b) must simultaneously be squares but am struggling to get further

    got further

    can 48(a-b)+7 be a perfect square?
    Last edited by moriman; October 9th 2011 at 02:02 AM.
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  2. #2
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    Re: can this expression be a perfect square?

    48(a-b)+7=x^2

    a and b are integers a > b > 0

    48a-48b+3=x^2-4

    48a-48b+3=(x+2)(x-2)

    3(16a-16b+1)=(x+2)(x-2)

    so\ \ \ \ x+2=3\ \ \ \ x=1

    or\ \ \ \ x-2=3\ \ \ \ x=5

    if\ \ x=1\ \ then\ \ 48(a-b)=-6\ \ \ =>\ \ \ a < b

    if\ \ x=5\ \ then\ \ 48(a-b)=18\ \ \ a-b=\dfrac{3}{8}\ \ \ \ =>\ \ \ \ a\ and\ b\ are\ not\ integers


    Is this correct
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Re: can this expression be a perfect square?

    Quote Originally Posted by moriman View Post
    48(a-b)+7=x^2

    a and b are integers a > b > 0

    48a-48b+3=x^2-4

    48a-48b+3=(x+2)(x-2)

    3(16a-16b+1)=(x+2)(x-2)

    so\ \ \ \ x+2=3\ \ \ \ x=1

    or\ \ \ \ x-2=3\ \ \ \ x=5

    if\ \ x=1\ \ then\ \ 48(a-b)=-6\ \ \ =>\ \ \ a < b

    if\ \ x=5\ \ then\ \ 48(a-b)=18\ \ \ a-b=\dfrac{3}{8}\ \ \ \ =>\ \ \ \ a\ and\ b\ are\ not\ integers


    Is this correct
    This looks right, but you could have just noted that if 48(a-b)+7=x^2 then 3 is a quadratic residue \text{mod }4, which it isn't.
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  4. #4
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    Re: can this expression be a perfect square?

    Thanks Drexel.

    I'll have to take a look at quadratic residues, it's a loooooooonnnnng time since I was at school
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