# A hard problem for me

• April 4th 2010, 07:40 AM
havaliza
A hard problem for me
if $b|a^2+1$ and $b>a>1$ prove that $b>a+\sqrt{a}$
• April 4th 2010, 12:18 PM
chisigma
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

$\chi$ $\sigma$
• April 4th 2010, 12:25 PM
tonio
Quote:

Originally Posted by havaliza
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
• April 5th 2010, 09:46 AM
havaliza
Quote:

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 $b^2-2ab+a^2>2b^2-2a$?
• April 5th 2010, 10:49 AM
chiph588@
Quote:

Originally Posted by havaliza
Can I ask why $b^2-2ab+a^2>2b^2-2a$?

Since $a^2>0$.
• April 7th 2010, 09:46 PM
simplependulum
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