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Math Help - Is there a relationship that can be pinned down?

  1. #1
    Junior Member
    Joined
    Aug 2011
    From
    Omagh
    Posts
    55

    Is there a relationship that can be pinned down?

    Back again,

    After alexmahone's and CaptainBlack's excellent work on a previous problem

    http://www.mathhelpforum.com/math-he...tml#post678257

    I've worked myself into another corner.

    Tackling this equation

    f^2=\left(\dfrac{2p^2+2pq+q^2}{2}-b\right)^2-2b^2

    with the same method as the previous post, this time using r and s as factors of b, I ended up with the result I was expecting but hoping against.

    2p^2+2pq+q^2=2r^2+2rs+s^2

    I want results other that the trivial r=p and s=q

    I don't think that
    r and s can be positive integers and satisfy this equation, but r^2, s^2 and rs certainly can.

    When p=7, q=4, r^2=72, s^2=2 and rs=12 then

    2p^2+2pq+q^2=2r^2+2rs+s^2 with r\not=p\ \ r\not=q\ \ s\not=p\ \ and\ \ s\not=q

    I tried using the quadratic formula, solving first for r, then substituting this value back in and solving for s.

    This only led to

    \left(\sqrt{s^2-2(s^2-2p^2-2pq-q^2 )}\right)s=\left(\sqrt{s^2-2(s^2-2p^2-2pq-q^2 )}\right)s

    Not much help then.

    I've tried different ways for getting the difference of squares

    i.e. 2r^2+2rs+s^2=(r-s)^2+\left(\sqrt{r^2+4rs}\right)^2

    to get (p+q)^2+p^2=(r-s)^2+\left(\sqrt{r^2+4rs}\right)^2

    but these lead to incorrect results.

    How can I calculate the relationship between p, q, r and s?

    Thanks

    Pro
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  2. #2
    Junior Member
    Joined
    Aug 2011
    From
    Omagh
    Posts
    55

    Re: Is there a relationship that can be pinned down?

    Heading the quadratic direction I have got to here..

    b=rs

    r^2=\dfrac{\sqrt{f^2+2b^2}+f}{2}

    s^2=\sqrt{f^2+2b^2}-f

    checking this back gives

    r^2s^2=\dfrac{\left(\sqrt{f^2+2b^2}+f\right)\left(  \sqrt{f^2+2b^2}-f\right)}{2}=\dfrac{\left(f^2+2b^2\right)-f^2}{2}=b^2

    so r^2s^2=b^2 which agrees with rs=b

    For r^2 and s^2 to be integers we need f^2+2b^2 to be a perfect square.

    let Y^2=f^2+2b^2

    Y^2-f^2=2b^2

    so that \left(\dfrac{Y+f}{2}\right)^2+\left(\dfrac{Y-f}{2}\right)^2=b^2

    let \left(\dfrac{Y+f}{2}\right)=m^2-n^2 and \ \ \left(\dfrac{Y+f}{2}\right)=2mn so \ \ b=m^2+n^2

    Y+f=2m^2-2n^2.....(1)
    Y-f=4mn............(2)

    (1)+(2)\ \ \ \ Y=m^2+2mn-n^2\ \ \ \ \ (=\sqrt{f^2+2b^2})

    (1)-(2)\ \ \ \ f=m^2-2mn-n^2

    substituting these give

    r^2=\dfrac{\sqrt{f^2+2b^2}+f}{2}=m^2-n^2

    s^2=\sqrt{f^2+2b^2}-f=4mn

    but multiplying these gives r^2s^2=4mn(m^2-n^2)

    whereas earlier we had r^2s^2=(m^2+n^2)^2

    Have I made a mistake or do I need to find an m and n where

    4mn(m^2-n^2)=(m^2+n^2)^2

    Thanks
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