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Thread: A hard problem for me

  1. #1
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    A 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 $\displaystyle b>a+\sqrt{a}$
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  2. #2
    MHF Contributor chisigma's Avatar
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    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

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
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    Quote Originally Posted by havaliza View Post
    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
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  4. #4
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    Quote Originally Posted by tonio View Post
    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 $\displaystyle b^2-2ab+a^2>2b^2-2a$?
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by havaliza View Post
    Can I ask why $\displaystyle b^2-2ab+a^2>2b^2-2a$?
    Since $\displaystyle a^2>0 $.
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  6. #6
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    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
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