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  • #22510
    space time
    Participant

    a and b are natural numbers

    (a^2 + b^2 -d)^n = ( a^n + b^n)^2

    n > 2 , d is an whole number whose value ranges from 0 to ab

    prove that no natural solutions exist

    #23205
    Cheenta Support
    Participant

    For \(n>3\), if \(n\) is odd, then this means \(a^n+b^n\) is a perfect \(n^{th}\) power. By Fermat’s Last Theorem,
    no positive integer solutions exist.

    If \(n\) is even, suppose that \(n=2k\). Then, let \(x=a^2\) and \(y=b^2\). The equation becomes:
    \((x+y-d)^k=x^k+y^k\). By Fermat’s Last Theorem, no positive integer solutions exist as well.

    We now need to check \(k=1,2\). For the first, note that \(a^2+b^2<(a+b)^2\), so no solutions.
    For the second,
    \(a^2+b^2-d=a^2+b^2\) implies \(d=0\)
    So any \(a,b\) work with \(d=0\). However, if the range of \(d\) is exclusive, there are no solutions.

    The solution is Provided by our teaching assistant Tarit Goswami.

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