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Math Help - floor equality

  1. #1
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    Thumbs up floor equality

    Hello, Thanks for your help
    Prove that :

    \forall n\in\mathbb{N}^{*},\;\lfloor\sqrt{n}+\sqrt{n+1}\rf  loor =\lfloor\sqrt{4n+2}\rfloor.
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  2. #2
    Moo
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    Hello,

    I found it somewhere


    First prove that \sqrt{n}+\sqrt{n+1}<\sqrt{4n+2}

    \begin{aligned}<br />
4n^2+4n<4n^2+4n+1 & \Leftrightarrow (2\sqrt{n(n+1)})^2<(2n+1)^2 \\<br />
& \Leftrightarrow 2\sqrt{n(n+1)}<2n+1 \\<br />
& \Leftrightarrow n+(n+1)+2 \sqrt{n} \sqrt{n+1}<4n+2 \\<br />
& \Leftrightarrow (\sqrt{n}+\sqrt{n+1})^2<4n+2 \\<br />
& \Leftrightarrow \sqrt{n}+\sqrt{n+1}<\sqrt{4n+2}<br />
\end{aligned}


    Then prove that there exists no integer k such that \sqrt{n}+\sqrt{n+1}<k\leq \sqrt{4n+2} (meaning that their integer part is the same)

    Square it :
    2n+1+2\sqrt{n(n+1)}<k^2 \leq 4n+2
    2n+1+2 \sqrt{n^2+n}<k^2\leq 4n+2

    From \sqrt{n^2+n}<\sqrt{n^2}=n, we can say that 2n+2n+1=4n+1<2n+1+2 \sqrt{n^2+n}

    Thus we finally have :
    4n+1<k^2\leq 4n+2

    Since k is an integer, k^2 is also an integer.
    Thus k^2=4n+2
    But a squared number can only be 0 or 1 modulo 4.
    Therefore, this integer k doesn't exist.
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  3. #3
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    hello
    thanks Moo

    why the squared number can only be 0 or 1 modulo 4 ? you can give me the proof
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  4. #4
    Moo
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    Quote Originally Posted by linda2005 View Post
    hello
    thanks Moo

    why the squared number can only be 0 or 1 modulo 4 ? you can give me the proof
    Here is, imho, the simplest way :
    a number can only be even or odd.
    - if it's even : n=2k --> nČ=4kČ, which is 0 modulo 4
    - if it's odd : n=2k+1 --> nČ=4kČ+4k+1, which is 1 modulo 4
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