Let $\displaystyle A\in\mathbb{N}>1$ and a natural divisor B such that $\displaystyle B|A^2+1$.If $\displaystyle B-A>0$, so proof that $\displaystyle B-A>\sqrt{A}$

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- Jun 30th 2009, 09:52 PMstreethotInequalitie
Let $\displaystyle A\in\mathbb{N}>1$ and a natural divisor B such that $\displaystyle B|A^2+1$.If $\displaystyle B-A>0$, so proof that $\displaystyle B-A>\sqrt{A}$

- Jul 1st 2009, 04:34 AMPaulRS
Just note that $\displaystyle B|[(B-A)^2+1]$ and so: $\displaystyle (B-A)^2+1\geq{B}$ then $\displaystyle B-A\geq{\sqrt{B-1}}$ (*)

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

1. $\displaystyle A<B$

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

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