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Thread: Equation manipulation

  1. #1
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    Equation manipulation

    I was hoping that some of you with a lot more experience than me would check my work below for any blatant errors or incorrect assumptions, I have marked the parts I'm unsure about. I have deliberately included every step for clarity.

    Many thanks in advance.

    Two equations:

    $\displaystyle b=u^2+4uv+2v^2$ ......... (1)

    $\displaystyle a^2+c^2=2(u^2+2uv+2v^2)^2$............ (2)

    Restrictions:

    u and v are integers $\displaystyle u>v\sqrt{2}>0$
    a, b and c are integers a > b > c > 0

    I am trying to prove, given the 2 equations and restrictions, that there are no valid values for a, b or c. Proof that any one of them can't be an integer within the restrictions is enough.

    From (2)

    $\displaystyle \left(\dfrac{a+c}{2}\right)^2+\left(\dfrac{a-c}{2}\right)^2=(u^2+2uv+2v^2)^2$

    manipulating the RHS (I'll call this Step 1 for a later question)

    $\displaystyle \left(\dfrac{a+c}{2}\right)^2+\left(\dfrac{a-c}{2}\right)^2=(u^2+2v^2 )^2+4uv(u^2+uv+2v^2)$

    $\displaystyle let\ \ \ \ \ \left(\dfrac{a+c}{2}\right)^2=(u^2+2v^2 )^2\ \ \ \ \ \ \ \ \ \ \ \ (3)$

    $\displaystyle and\ \ \ \ \left(\dfrac{a-c}{2}\right)^2=4uv(u^2+uv+2v^2)\ \ \ \ \ \ \ (4)$

    everything ok so far? (I think the min ratio of u:v has increased here, but not important at the minute)

    from (3) .... $\displaystyle (a+c)^2=4(u^2+2v^2 )^2\ \ \ \ \ \ \ \ (5)$

    $\displaystyle a+c=2(u^2+2v^2 )\ \ \ \ \ \ \ \ (6)$

    from (4) .... $\displaystyle (a-c)^2=16uv(u^2+uv+2v^2)$

    manipulating LHS

    $\displaystyle (a+c)^2-4ac=16uv(u^2+uv+2v^2)\ \ \ \ \ \ \ (7)$

    $\displaystyle (5)-(7)\ \ \ 4ac=4(u^4-4u^3v-8uv^3+4v^4)$

    $\displaystyle ac=u^4-4u^3v-8uv^3+4v^4\ \ \ \ \ \ \ \ (8)$

    from (6) .... $\displaystyle c=2(u^2+2v^2 )-a$

    substituting this in (8) ...

    $\displaystyle (2(u^2+2v^2 )-a)a=u^4-4u^3v-8uv^3+4v^4$

    $\displaystyle a^2-2(u^2+2v^2 )a+(u^4-4u^3v-8uv^3+4v^4)=0$

    $\displaystyle a=(u^2+2v^2 )+2\sqrt{u^3v+u^2v^2+2uv^3}$

    for $\displaystyle a$ to be +ve integer $\displaystyle u^3v+u^2v^2+2uv^3$ must be a perfect square.

    $\displaystyle let\ \ \ z^2=u^3v+u^2v^2+2uv^3$

    no manipulation here, just grouping...

    $\displaystyle z^2=(uv)^2+(u^2+2v^2)uv$

    $\displaystyle (z+uv)(z-uv)=(u^2+2v^2)uv$

    $\displaystyle so\ either\ z+uv=uv\ \ \ \ \ \ z=0$

    $\displaystyle or\ \ z-uv=uv\ \ \ \ \ z=2uv$

    are these two roots attained correctly?

    $\displaystyle z=0\implies\ a=u^2+2v^2\implies\ (from\ (6))\ c=u^2+2v^2\ \ \therefore\ a=c\ so\ z\not=0$

    $\displaystyle z=2uv\implies\ a=(u^2+2v^2 )+4uv\ \ \therefore\ a=b\ so\ z\not=2uv$

    So this proves that $\displaystyle a$ cannot have a value that fits within the given constraints.

    Question about Step 1. Does what I have shown only prove there's no valid $\displaystyle a$ when the equations are manipulated as in Step 1, or does it prove it conclusively?

    If not conclusively, there are infinite ways of manipulating the equation at that stage $\displaystyle \left(i.e.\ \ \ (u^{100}+v^{250})^2+whatever\right)$

    If this is the case, is there any way to prove conclusively what I'm attempting?

    Many thanks for your time
    Last edited by moriman; Oct 10th 2011 at 03:09 PM.
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  2. #2
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    Re: Equation manipulation

    your equation (2) ...

    $\displaystyle a+c = 2(u^2 + 2uv + 2v^2)^2$
    divide both sides by 2 ...

    $\displaystyle \frac{a+c}{2} = (u^2 + 2uv + 2v^2)^2$


    ... now, how did this next equation come about?

    From (2)

    $\displaystyle \left(\frac{a+c}{2}\right)^2 + \left(\frac{a-c}{2}\right)^2 = (u^2 + 2uv + 2v^2)^2$
    are you saying $\displaystyle \frac{a+c}{2} = \left(\frac{a+c}{2}\right)^2 + \left(\frac{a-c}{2}\right)^2$ ?
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  3. #3
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    Re: Equation manipulation

    sigh, sorry, I checked and rechecked the op and missed that typo

    I have edited the 2nd equation to the correct

    $\displaystyle a^2+c^2=2(u^2+2uv+2v^2)^2$............ (2)

    sorry again
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