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Thread: [SOLVED] Dividing problem

  1. #1
    Moo
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    [SOLVED] Dividing problem

    Hi !

    I've got a little problem, seems simple... Here it is :


    Text :
    We're going to study all couples (n,k) such as :

    $\displaystyle {n \choose k}={n+1 \choose k-1}$


    Question :
    Let $\displaystyle p=n-k+1$ and $\displaystyle t=gcd(p,k)$. Also $\displaystyle s=\frac{k}{t}$

    Show that $\displaystyle p=t^2$


    So far... :
    By replacing the combinations with their formulas with !, i've found that :

    $\displaystyle p(p+1)=(p+k)k$

    $\displaystyle p^2+p=pk+k^2$

    So it's easy to show that $\displaystyle t^2|p$


    And now... ?
    I can't find how to show that $\displaystyle p|t^2$ or $\displaystyle p \leq t^2$ :'(

    I thought Bézout's identity could help, but no...


    Thanks in advance !
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Moo View Post
    I've found that :

    $\displaystyle p(p+1)=(p+k)k$
    $\displaystyle p^2+p=pk+k^2$

    So it's easy to show that $\displaystyle t^2|p$


    And now... ?
    I can't find how to show that $\displaystyle p|t^2$ or $\displaystyle p \leq t^2$ :'(
    So if $\displaystyle t = \gcd(p,k)$, let p = rt and k = st, where r and s are co-prime.

    Then $\displaystyle rt(rt+1) = t^2s(r+s)$, hence $\displaystyle r(rt+1) = s(r+s)t$.

    Since t has to divide the left-hand side of the last equation, and t is clearly co-prime with rt+1, it follows that t divides r (so that t^2 divides p: you already knew that).

    Similarly, since r has to divide the right-hand side of that equation, and r is co-prime with both s and r+s, it follows that r divides t.

    Put those two facts together, and you have r=t as required.
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  3. #3
    Moo
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    Hello,

    Thanks for the answer ! This is exactly what i wanted (and even for the following question !)
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