# Math Help - Inequalitie

1. ## 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 $B-A>\sqrt{A}$

2. 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 (**) because:

1. $A
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 (*).