Please help me to solve this problem. I've worked on it about a week or two. But it's very hard for me:if $\displaystyle b|a^2+1$ and $\displaystyle b>a>1$ prove that $\displaystyle b>a+\sqrt{a}$
I suppose $\displaystyle a,b\in\mathbb{N}$ , so assume $\displaystyle \sqrt{a}\geq b-a\Longrightarrow a\geq b^2-2ab+a^2>2b^2-2a=2a(a-1)\Longrightarrow$ $\displaystyle 1>2(a-1)\Longrightarrow a<\frac{3}{2}$ , which is a contradiction
to $\displaystyle a>1\,,\,\,a\in\mathbb{N}$ , so we're done.
Tonio
another method :
If $\displaystyle a < b \leq a + \sqrt{a} $ , let
$\displaystyle b = a + k ~~, 0 < k \leq \sqrt{a} $
$\displaystyle a^2 + 1 = 0 mod(a + k) ~\implies k^2+ 1 = 0 mod(a+k) $
since $\displaystyle k^2 + 1 \neq 0 $ , $\displaystyle a+k \leq k^2 + 1 $
but $\displaystyle k^2 + 1 \leq \sqrt{a}^2 + 1 = a+1 $ we have
$\displaystyle a+1 \geq a + k ~\implies k=1 ~\implies a=1 $ , a contradiction