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$\displaystyle b>a+\sqrt{a}$

Printable View

- Apr 4th 2010, 07:40 AMhavalizaA hard problem for me
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**$\displaystyle b>a+\sqrt{a}$ - Apr 4th 2010, 12:18 PMchisigma
If $\displaystyle a>1$ then is $\displaystyle a^{2}+1 > a + \sqrt{a}$, so that if $\displaystyle b|a^{2}+1$ is $\displaystyle b\ge a^{2}+1$ and then $\displaystyle b>a + \sqrt{a}$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Apr 4th 2010, 12:25 PMtonio

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 - Apr 5th 2010, 09:46 AMhavaliza
- Apr 5th 2010, 10:49 AMchiph588@
- Apr 7th 2010, 09:46 PMsimplependulum
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