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Math Help - The ceiling of the square root of n problem.

  1. #1
    Junior Member
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    Red face The ceiling of the square root of n problem.

    How can one find all integers n for which \lceil <br />
\sqrt{n}\rceil divides n?
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  2. #2
    Member
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    Nov 2009
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    Let n \in \mathbb{N}, and let \lceil \sqrt{n} \rceil = a. Clearly, (a-1)^2 < n \le a^2. Furthermore, the only numbers between (a-1)^2 and a^2 which a divides are a^2 and a^2-a ( a^2-2a is already outside the interval, as a^2-2a<a^2-2a+1=(a-1)^2). Thus, \lceil \sqrt{n} \rceil \mid n \Leftrightarrow n=a^2-a\;\:\text{or}\:\;n=a^2,\:\;1<a \in \mathbb{N}

    The last step is not fully explained, and I'll let you complete the gap.
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