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Math Help - 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 b|a^2+1 and b>a>1 prove that b>a+\sqrt{a}
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  2. #2
    MHF Contributor chisigma's Avatar
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    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
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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 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
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  4. #4
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    Quote Originally Posted by tonio View Post
    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?
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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 b^2-2ab+a^2>2b^2-2a?
    Since  a^2>0 .
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  6. #6
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    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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