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Math Help - even more a+b=2c

  1. #1
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    Omagh
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    even more a+b=2c

    Back again

    For those who didn't see earlier thread I'll just put this first

    a^2+b^2=2c^2

    we can write a, b & c like this:
     a=m^2+2m-1\ \ \ \ (1)
     b=m^2-2m-1\ \ \ \ (2)
     c=m^2+1\ \ \ \ \ \ \ \ \ \ \ \ (3)
    Proof:

    a^2=m^4+4m^3+2m^2-4m+1
    b^2=m^4-4m^3+2m^2+4m+1
    c^2=m^4+2m^2+1

    so\ \ \ a^2+b^2=2m^4+4m^2+2\ \ \ =2c^2\ \ \ Q.E.D.
    Now\ \ \ 194^2+142^2=2(170^2)

    let a=194, b=142 and c=170

    using (1)-(2) we have \ \ a-b=4m\ \ \4m=194-142=52\ \ \ \ so\ m=13

    substituting this in (1), (2) & (3) gives the expected results

    (1)\ \ \ a=m^2+2m-1\ \ \ 169+26-1=194
    (2)\ \ \ b=m^2-2m-1\ \ \ 169-26-1=142
    (3)\ \ \ c=m^2+1\ \ \ \ \ \ \ \ \ \ \ 169+1=170

    All good so far but...

    527^2+289^2=2(425^2)

    let a=527, b=289, c=425

    using (1)-(2) we have \ \ a-b=4m\ \ \4m=527-289=238\ \ \ \ so\ m=59.5

    substituting this in (1), (2) & (3) gives

    (1)\ \ \ a=m^2+2m-1\ \ \ 3540.25+119-1=3658.25\ \ \ \ not\ the\ expected\  527
    (2)\ \ \ b=m^2-2m-1\ \ \ 3540.25-119-1=3420.25\ \ \ \ not\ the\ expected\  289
    (3)\ \ \ c=m^2+1\ \ \ \ \ \ \ \ \ 3540.25+1=3541.25\ \ \ \ not\ the\ expected\  425

    What am I doing wrong or misunderstanding?

    Many thanks

    Pro
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  2. #2
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    Re: even more a+b=2c

    Ignore this post.

    Added in Edit:

    Pro,

    What I meant was: "Ignore my post." I had written something that was incorrect & didn't see how to delete my post.
    Last edited by SammyS; August 26th 2011 at 01:35 PM. Reason: Brain Cramp.
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  3. #3
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    Omagh
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    Re: even more a+b=2c

    Quote Originally Posted by SammyS View Post
    Ignore this post.
    [edit]

    ah! just noticed your reason for editing

    No problem SammyS, lol

    Thanks

    Pro
    Last edited by procyon; August 26th 2011 at 01:40 PM. Reason: can't bloody read, lol
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  4. #4
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    Re: even more a+b=2c

    Quote Originally Posted by procyon View Post
    Back again

    For those who didn't see earlier thread I'll just put this first

    a^2+b^2=2c^2

    we can write a, b & c like this:
     a=m^2+2m-1\ \ \ \ (1)
     b=m^2-2m-1\ \ \ \ (2)
     c=m^2+1\ \ \ \ \ \ \ \ \ \ \ \ (3)
    Proof:

    a^2=m^4+4m^3+2m^2-4m+1
    b^2=m^4-4m^3+2m^2+4m+1
    c^2=m^4+2m^2+1

    so\ \ \ a^2+b^2=2m^4+4m^2+2\ \ \ =2c^2\ \ \ Q.E.D.
    Now\ \ \ 194^2+142^2=2(170^2)

    let a=194, b=142 and c=170

    using (1)-(2) we have \ \ a-b=4m\ \ \4m=194-142=52\ \ \ \ so\ m=13

    substituting this in (1), (2) & (3) gives the expected results

    (1)\ \ \ a=m^2+2m-1\ \ \ 169+26-1=194
    (2)\ \ \ b=m^2-2m-1\ \ \ 169-26-1=142
    (3)\ \ \ c=m^2+1\ \ \ \ \ \ \ \ \ \ \ 169+1=170

    All good so far but...

    527^2+289^2=2(425^2)

    let a=527, b=289, c=425

    using (1)-(2) we have \ \ a-b=4m\ \ \4m=527-289=238\ \ \ \ so\ m=59.5

    substituting this in (1), (2) & (3) gives

    (1)\ \ \ a=m^2+2m-1\ \ \ 3540.25+119-1=3658.25\ \ \ \ not\ the\ expected\  527
    (2)\ \ \ b=m^2-2m-1\ \ \ 3540.25-119-1=3420.25\ \ \ \ not\ the\ expected\  289
    (3)\ \ \ c=m^2+1\ \ \ \ \ \ \ \ \ 3540.25+1=3541.25\ \ \ \ not\ the\ expected\  425

    What am I doing wrong or misunderstanding?

    Many thanks

    Pro
    Once you choose any one of a, or b, or c, then you have determined the value for m.

    Your first example works because m=13 for a = 194, b = 142, and c = 170.

    For your second case, if a = 527, then m = 22, so b = 439 c = 485.

    If you want b = 289, then m=1+\sqrt{291}, so a=293+4\sqrt{291}\,, and c=293+2\sqrt{291}\,,

    ... etc.

    You can pick an integer, m, then get a, b, and c, so that a^2+b^2=2c^2\,, but simply finding a, b, c that satisfy a^2+b^2=2c^2 does not ensure that there is a number m that generates a, b, and c.
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  5. #5
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    Omagh
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    [solved] Re: even more a+b=2c

    You can pick an integer, m, then get a, b, and c, so that \ \ a^2+b^2=2c^2 but simply finding a, b, c that satisfy \ \ a^2+b^2=2c^2 does not ensure that there is a number m that generates a, b, and c.
    I've come to the same conclusion but still can't really get my head around it
    It just looks so right.

    The m I gave wasn't just for a=527. It was generated using (1) and (2) as simultaneous equations, but the result doesn't satisfy (1) and (2) simultaneously

    Ah! Because the equations aren't simultaneously correct!

    (1)-(2) => m=59.5
    (1)+(2) => m=sqrt(409)
    Last edited by procyon; August 26th 2011 at 04:15 PM. Reason: forgot to add
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