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Re: [PrimeNumbers] Binomial and prime congruence

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  • Phil Carmody
    ... It s a 3-line induction proof. ... Not that I know of. I know that the power of each prime that divides a binomial expression is one that has been
    Message 1 of 2 , Jul 30, 2006
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      --- Sebastian Martin <sebi_sebi@...> wrote:
      > Hello all:
      >
      > Can anyone prove the following congruence?
      >
      > Binomial(m,p)=Floor[m/p] (mod p)
      >
      > for all m integer and for all p prime number.

      It's a 3-line induction proof.

      > Corollary:
      >
      > If pn and p(n+1) are consecutive pime numbers
      >
      > Binomial[p(n+1),pn]=1 (mod pn)
      >
      > (it is easy, consider Bertrand postulate)
      >
      > There are any previous reference?

      Not that I know of. I know that the power of each prime that divides a binomial
      expression is one that has been mentioned in the past. The expressions
      discovered are useful for fast computation of binomials.

      Phil

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