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Math Help - Sum of binomials

  1. #1
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    Sum of binomials

    Let p>3 be a prime. \sum_{i=1}^p\binom {i\cdot p}{p}\cdot\binom {\left(p-i+1\right)p}{p}\equiv ?\mod p^2.
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  2. #2
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    Quote Originally Posted by james_bond View Post
    Let p>3 be a prime. \sum_{i=1}^p\binom {i\cdot p}{p}\cdot\binom {\left(p-i+1\right)p}{p}\equiv ?\mod p^2.
    the answer is \frac{p(p+1)(p+2)}{6} and we don't need the condition p > 3. to prove this let f(x)=\prod_{k=1}^{p-1}(x-k)=x^{p-1}-a_1x^{p-2} + \cdots -a_{p-2}x + a_{p-1}.

    by the Fermat's little theorem, in (\mathbb{Z}/p\mathbb{Z})[x], we have f(x)=x^{p-1}-1. as a result (p-1)!=a_{p-1} \equiv -1 \mod p, which is Wilson's theorem, and

    a_j \equiv 0 \mod p, for all j \neq p-1. thus, for any n \in \mathbb{N}: \ \binom{np}{p}=\frac{np(np-1) \cdots (np-p+1)}{p!}=\frac{nf(np)}{(p-1)!} \equiv n \mod p^2. so your sum, modulo p^2,

    is equal to \sum_{i=1}^{p}i(p-i+1)=\frac{p(p+1)(p+2)}{6}. \ \Box


    Remark 1. this method gives another proof for the problem you asked in here.

    Remark 2. there's a theorem (i don't remember its name) which says that for any prime number p > 3: \ a_{p-2} \equiv 0 \mod p^2. does anybody know the

    name of this theorem? anyway, using this result we get this stronger result that, modulo p^3, your sum for p > 3 is still equal to \frac{p(p+1)(p+2)}{6}.
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    Quote Originally Posted by NonCommAlg View Post

    Remark 2. there's a theorem (i don't remember its name) which says that for any prime number p > 3: \ a_{p-2} \equiv 0 \mod p^2. does anybody know the name of this theorem?
    ok, i found it! it's Wolstenholme's theorem.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    ok, i found it! it's Wolstenholme's theorem.
    I can't imagine that's a particularly commonly used theorem.
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    I can't imagine that's a particularly commonly used theorem.
    well, it's a non-trivial result and that's good enough to make it useful!
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