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Math Help - solution please...can anyone show the proof...

  1. #1
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    Question solution please...can anyone show the proof...

    Is there anyone who can show that the equation:

    n^4 - n^2*m^2+m^4 = s^2, (where m, n and s are positive integers) has no integer solutions with |n| > |m| except (n^2, m^2) = (1, 0)


    I will appreciate it if anyone can give at least a reference where the proof can be found.

    Thanks!
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  2. #2
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    I am working on this.
    I see two ways to approach it, one method is below using Pythagorean Triples.
    ---
    You can state that,
    n^2-m^2=2uv
    nm=u^2-v^2
    s=u^2+v^2
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  3. #3
    Super Member malaygoel's Avatar
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    Quote Originally Posted by sheryl
    Is there anyone who can show that the equation:

    n^4 - n^2*m^2+m^4 = s^2, (where m, n and s are positive integers) has no integer solutions with |n| > |m| except (n^2, m^2) = (1, 0)


    I will appreciate it if anyone can give at least a reference where the proof can be found.

    Thanks!
    Just a thought.

    We have to prove that there does not exist two integers n and m such that
    n^4 -n^2m^2 +m^4 is a perfect square.
    So we equalled it to s^2.
    We can solve the equation for n^2 and show that the discriminant cannot be zero.
    n^4 - m^2n^2 +(m^4-s^2)=0
    We get
    (n^2 - \frac{m^2\pm \sqrt{(m^2)^2-4(m^4-s^2)}}{2})=0
    (n^2-\frac{m^2 \pm \sqrt{4s^2 -3m^4}}{2})=0
    Since the expression is a perfect square, 4s^2 -3m^4 has to be zero which is not possible.
    We have also the case in which 4s^2 -3m^4 is a pefect square.

    Malay
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