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Thread: Modulo

  1. December 23rd 2008, 12:51 AM #1
    Sea
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    Modulo

    <br /> <br /> a,n\in\mathbb{Z},


    Show that:

    n+1=a^2 \Rightarrow n+1|2(n-1)!
    Last edited by Sea; December 23rd 2008 at 03:07 AM.
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  2. December 23rd 2008, 04:04 AM #2
    PaulRS
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    For the hypothesis it'd be enough to say that n is a positive integer such that n+1 is not a prime number.

    Let p^s be the maximum power of a prime p dividing n+1.


    • If n+1 is divisible by at least 2 primes, then p^s\leq{\tfrac{n}{2}}\leq{n-1} and so p^s appears as a factor of (n-1)!=1\cdot{2}...\cdot{p^s}\cdot{...\cdot{(n-1)}}, thus p^s|(n-1)!
    • Otherwise n+1=p^s. If p=2 then 2^{s-1} appears as a factor of (n-1)! - this can be seen as in the previous part- so 2^{s-1}|(n-1)! and so 2^{s}|[2\cdot{(n-1)!}]. Now if p>2 we have that p^{s-1} is a factor of (n-1)!, but, since n is not prime, it must be that s>1 and so 2p appears also as a factor in the product, because 2p<p^2-1\leq{p^s-1}=n-1 (*). Clearly 2p\neq p^{s-1} so we have that p^s|(n-1)!

    (*) 2p<p^2-1 iff 2<p^2-2p+1=(p-1)^2 and this holds since we were considering p>2

    From this analysis it follows that p^s|[2\cdot{(n-1)!}]. (this holds for all the maximum powers of primes dividing n+1)

    And therefore (n+1)|[2\cdot{(n-1)!}]

    Remark: This doesn't hold when n+1 is a prime greater than 2. ( it holds for n+1=2)
    Last edited by PaulRS; December 23rd 2008 at 08:49 AM. Reason: the case n+1=2 was missing in the remark
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