# A hard problem for me

• Apr 4th 2010, 07:40 AM
havaliza
A hard problem 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 PM
chisigma
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 PM
tonio
Quote:

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

Originally Posted by tonio
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

Can I ask why $\displaystyle b^2-2ab+a^2>2b^2-2a$?
• Apr 5th 2010, 10:49 AM
chiph588@
Quote:

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

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