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How would you go about proving that $$ {p \choose k} \equiv 0 \pmod p $$ if $0 < k < p$ and $p$ is prime?

Edit: I'm fairly certain that I should begin at the definition of the binomial coefficient, namely that $$ {p \choose k} = \frac{p!}{(p - k)! \cdot k!} $$ and rewrite it as $$ \frac{p \cdot (p - 1)!}{(p - k)! \cdot k!} = p \cdot \frac{(p - 1)!}{(p - k)! \cdot k!} $$ which shows that it is a multiple of $p$ and thus equivalent to zero modulo $p$. However, I know that this is not complete and I am unsure of what to do next.

dfeuer
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Matthew Turner
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1 Answers1

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HINT: $$\binom{p}k=\frac{p!}{k!(p-k)!}\;;$$ if $0<k<p$, how many factors of $p$ are there in the denominator? Remember, $p$ is prime.

Brian M. Scott
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    I don't see what you mean... – Matthew Turner Oct 16 '13 at 06:09
  • @Matthew: There’s clearly a factor of $p$ in the numerator; you found that yourself. Look at $k!$: it’s $k(k-1)(k-2)\ldots(2)(1)$. Is any of the numbers $k,k-1,\ldots,2,1$ divisible by $p$? – Brian M. Scott Oct 16 '13 at 06:11
  • No, none of them are divisible by $p$ because $k < p$. – Matthew Turner Oct 16 '13 at 06:22
  • @Matthew: and because $k>0$, no factor of $(n-k)!$ is divisible by $p$. So can the denominator contain any factors of $p$ at all$? – Brian M. Scott Oct 16 '13 at 06:23
  • No? I don't think so... – Matthew Turner Oct 16 '13 at 06:40
  • @Matthew: That’s right: it can’t. If $p$ were composite it could: you could get a factor of $6$, for instance, from $2!4!$. But when $p$ is prime, that can’t happen: if it’s a factor of a product, it must divide one of the factors forming that product. $k!(n-k)!$ has no factor divisible by $p$, so is isn’t divisible by $p$ either. – Brian M. Scott Oct 16 '13 at 06:47