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Thread: 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

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

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

    $\displaystyle 2p^2+2pq+q^2=2r^2+2rs+s^2$

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

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

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

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

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

    This only led to

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

    Not much help then.

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

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

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

    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..

    $\displaystyle b=rs$

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

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

    checking this back gives

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

    so $\displaystyle r^2s^2=b^2$ which agrees with $\displaystyle rs=b$

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

    let $\displaystyle Y^2=f^2+2b^2$

    $\displaystyle Y^2-f^2=2b^2$

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

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

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

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

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

    substituting these give

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

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

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

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

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

    $\displaystyle 4mn(m^2-n^2)=(m^2+n^2)^2$

    Thanks
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