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Thread: proving two expressions can be squares

  1. #1
    Apr 2012

    proving two expressions can be squares


    If it is known that $\displaystyle (A^2+B^2)(C^2+D^2)$ is a perfect square
    where A, B, C and D are integers, A and B are coprime, C and D are coprime and D > C >= B > A > 0

    $\displaystyle (A^2+B^2)(C^2+D^2)+4(A^2-B^2)CD$ can be a perfect square..........(1)
    and $\displaystyle \ (A^2+B^2)(C^2+D^2)-4(A^2-B^2)CD$ can be a perfect square..........(2)

    but can it be proven that (1) and (2) can't be perfect squares simultaneously (or examples to the contrary) ?


    A = 1, B = 2, C = 2, D = 11

    $\displaystyle (A^2+B^2)(C^2+D^2)=25^2$
    $\displaystyle (A^2+B^2)(C^2+D^2)+4(A^2-B^2)CD=19^2$
    $\displaystyle \ (A^2+B^2)(C^2+D^2)-4(A^2-B^2)CD$ is not a perfect square

    A = 1, B = 2, C = 58, D = 209

    $\displaystyle (A^2+B^2)(C^2+D^2)=485^2$
    $\displaystyle (A^2+B^2)(C^2+D^2)+4(A^2-B^2)CD$ is not a perfect square
    $\displaystyle \ (A^2+B^2)(C^2+D^2)-4(A^2-B^2)CD=617^2$

    thank you

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  2. #2
    Senior Member
    Apr 2009
    Atlanta, GA

    Re: proving two expressions can be squares

    Well, recognize that what you want to construct is an arithmetic progression of three squares, where the middle square is $\displaystyle (A^2+B^2)(C^2+D^2)$ and the common difference is $\displaystyle 4(B^2-A^2)CD$ (reversing the order to indicate $\displaystyle B > A$). There is a parametric formula for progressions of three squares that says: $\displaystyle X^2,Y^2,Z^2 = (a^2-2ab-b^2)^2,(a^2+b^2)^2,(-a^2-2ab+b^2)^2$ for $\displaystyle a<b$, giving a common difference of $\displaystyle a^2-2ab-b^2$. (Click here for a wonderful proof that this captures all of them.) So, your question becomes:

    Find positive integers $\displaystyle A,B,C,D,a,b$ with $\displaystyle (A,B)=(C,D)=1, A<B\le C<D$ satisfying
    $\displaystyle (A^2+B^2)(C^2+D^2)=(a^2+b^2)^2$ and $\displaystyle 4(B^2-A^2)CD=a^2-2ab-b^2$

    You might start by isolating $\displaystyle a$ in the first equation, subbing it into the second, and attempting to isolate $\displaystyle b$. Not sure if this is possible directly because this gives you a quartic polynomial on $\displaystyle b$, but if you can, then the problem will reduce to proving when or whether your expression $\displaystyle b=f(A,B,C,D)$ evaluates to an integer. I know I'm being lazy, but that should give you something to chew on.
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  3. #3
    Super Member
    Jun 2012

    Re: proving two expressions can be squares

    Assume that they are both perfect squares, i.e.

    $\displaystyle (A^2 + B^2)(C^2 + D^2) + 4(A^2 - B^2)CD = k^2$
    $\displaystyle (A^2 + B^2)(C^2 + D^2) - 4(A^2 - B^2)CD = m^2$.

    Note the constraint, 0 < A < B, so $\displaystyle A^2 - B^2 < 0$. Taking their difference (and switching order a bit),

    $\displaystyle m^2 - k^2 = 8(B^2 - A^2)CD$

    What happens if 8CD is a perfect square? For example, try C = 32, D = 49. I'll let you play around with that.
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