A hard problem for me

**havaliza** · April 4th 2010, 07:40 AM

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 $b|a^2+1$ and $b>a>1$ prove that $b>a+\sqrt{a}$

**chisigma** · April 4th 2010, 12:18 PM

If $a>1$ then is $a^{2}+1 > a + \sqrt{a}$ , so that if $b|a^{2}+1$ is $b\ge a^{2}+1$ and then $b>a + \sqrt{a}$ ...

Kind regards

$\chi$ $\sigma$

**tonio** · April 4th 2010, 12:25 PM

Originally Posted by havaliza

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 $b|a^2+1$ and $b>a>1$ prove that $b>a+\sqrt{a}$

I suppose $a,b\in\mathbb{N}$ , so assume $\sqrt{a}\geq b-a\Longrightarrow a\geq b^2-2ab+a^2>2b^2-2a=2a(a-1)\Longrightarrow$ $1>2(a-1)\Longrightarrow a<\frac{3}{2}$ , which is a contradiction

to $a>1\,,\,\,a\in\mathbb{N}$ , so we're done.

Tonio

**havaliza** · April 5th 2010, 09:46 AM

Originally Posted by tonio

I suppose $a,b\in\mathbb{N}$ , so assume $\sqrt{a}\geq b-a\Longrightarrow a\geq b^2-2ab+a^2>2b^2-2a=2a(a-1)\Longrightarrow$ $1>2(a-1)\Longrightarrow a<\frac{3}{2}$ , which is a contradiction

to $a>1\,,\,\,a\in\mathbb{N}$ , so we're done.

Tonio

Can I ask why $b^2-2ab+a^2>2b^2-2a$ ?

**chiph588@** · April 5th 2010, 10:49 AM

Originally Posted by havaliza

Can I ask why $b^2-2ab+a^2>2b^2-2a$ ?

Since $a^2>0$ .

**simplependulum** · April 7th 2010, 09:46 PM

another method :

If $a < b \leq a + \sqrt{a}$ , let

$b = a + k ~~, 0 < k \leq \sqrt{a}$

$a^2 + 1 = 0 mod(a + k) ~\implies k^2+ 1 = 0 mod(a+k)$

since $k^2 + 1 \neq 0$ , $a+k \leq k^2 + 1$

but $k^2 + 1 \leq \sqrt{a}^2 + 1 = a+1$ we have

$a+1 \geq a + k ~\implies k=1 ~\implies a=1$ , a contradiction

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