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Math Help - 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:

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

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

    Restrictions:

    u and v are integers 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)

    \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)

    \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)

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

    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) .... (a+c)^2=4(u^2+2v^2 )^2\ \ \ \ \ \ \ \ (5)

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

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

    manipulating LHS

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

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

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

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

    substituting this in (8) ...

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

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

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

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

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

    no manipulation here, just grouping...

    z^2=(uv)^2+(u^2+2v^2)uv

    (z+uv)(z-uv)=(u^2+2v^2)uv

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

    or\ \ z-uv=uv\ \ \ \ \ z=2uv

    are these two roots attained correctly?

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

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

    So this proves that 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 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 \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; October 10th 2011 at 03:09 PM.
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  2. #2
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    Re: Equation manipulation

    your equation (2) ...

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

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


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

    From (2)

    \left(\frac{a+c}{2}\right)^2 + \left(\frac{a-c}{2}\right)^2 = (u^2 + 2uv + 2v^2)^2
    are you saying \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

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

    sorry again
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