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Math Help - Inequalitie

  1. #1
    Junior Member
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    Inequalitie

    Let A\in\mathbb{N}>1 and a natural divisor B such that B|A^2+1.If B-A>0, so proof that B-A>\sqrt{A}
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  2. #2
    Super Member PaulRS's Avatar
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    Just note that B|[(B-A)^2+1] and so: (B-A)^2+1\geq{B} then B-A\geq{\sqrt{B-1}} (*)

    But we must have A<B-1 (**) because:

    1. A<B
    2. A^2\equiv{-1}(\bmod.B)

    Suppose it was A=B-1, then we would have A^2\equiv{(-1)^2=1}(\bmod.B) a contradiction - unless B=2 or B=1, but this doesn't happen since B-A>0 and A\geq{2}-, thus (**) holds and we are done by (*).
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