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Math Help - show that the equation has only one solution

  1. #1
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    show that the equation has only one solution

    This question really has me stumped.....

    a^2 + b^2 = 6( c^2 + d^2 )

    Any clues or hints to start this question...

    I think we suppose to use congruences to try an solve this...

    Any help greatly appreciated
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  2. #2
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    Quote Originally Posted by Dreamer78692 View Post
    This question really has me stumped.....

    a^2 + b^2 = 6( c^2 + d^2 )

    Any clues or hints to start this question...

    I think we suppose to use congruences to try an solve this...

    Any help greatly appreciated

    The RHS is being expressed as the sum of two squares. This is possible iff all the primes =3\!\!\pmod 4 that divide the RHS

    appear at an even power....now, we have there 6=2\cdot 3 , so....try to complete the argument now.

    Tonio
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  3. #3
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    Sorry Im not quite getting it .....
    could you elaborate please...

    from what i read....
    c^2 + d^2 can be expressed as (2^k)*p1..pk*q1...ql
    a^2 + b^2 can be expressed as (2^g)*s1..sm*t1...tn

    where p1...pk and s1...sm are primes =1 mod 4
    and where q1...ql and t1...tn are primes =3 mod 4 with even power...

    then
    a^2 + b^2 = 6( c^2 + d^2 )
    can be expressed as
    (2^g)*s1..sm*t1...tn = 6*(2^k)*p1..pk*q1...ql

    am i on the right track... I don't really know where to go from there


    Another guy posted this solution is this correct
    MathBin.net - Untitled
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  4. #4
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    Quote Originally Posted by Dreamer78692 View Post
    Sorry Im not quite getting it .....
    could you elaborate please...

    from what i read....
    c^2 + d^2 can be expressed as (2^k)*p1..pk*q1...ql
    a^2 + b^2 can be expressed as (2^g)*s1..sm*t1...tn

    where p1...pk and s1...sm are primes =1 mod 4
    and where q1...ql and t1...tn are primes =3 mod 4 with even power...

    then
    a^2 + b^2 = 6( c^2 + d^2 )
    can be expressed as
    (2^g)*s1..sm*t1...tn = 6*(2^k)*p1..pk*q1...ql

    am i on the right track... I don't really know where to go from there


    Another guy posted this solution is this correct
    MathBin.net - Untitled


    I don't think it is: he only managed to prove, perhaps, that both a,b are multiples of 3...

    Tonio
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  5. #5
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    Quote Originally Posted by Dreamer78692 View Post
    then
    a^2 + b^2 = 6( c^2 + d^2 )
    can be expressed as
    (2^g)*s1..sm*t1...tn = 6*(2^k)*p1..pk*q1...ql

    am i on the right track... I don't really know where to go from there
    could you show me how to do it... pretty please
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  6. #6
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    Quote Originally Posted by Dreamer78692 View Post
    could you show me how to do it... pretty please

    The only thing I can really think of is what I already told you: it is a very well known result in number theory that the only primes that can

    be expressed as the sum of two squares are 2 \,\,and\,\,those\,\,which\,\,are\,\, =1\!\!\pmod 4 , and from here we get that a natural

    number is expressable as the sum of two squares iff all its prime divisors which are =3\!\!\pmod 4 appear in the prime decomposition

    of this natural number at even powers.

    Thus, for example, 5\cdot 7\cdot 13 can't be written as the sum of two squares, but 5\cdot 7^2\cdot 13 can...

    Tonio
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