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Let $p$ be prime and $k\in\mathbb N, \ 0 < k < p.$

Claim: $p | \begin{pmatrix}p \\ k \end{pmatrix}$

Proof:

We know that $\begin{pmatrix}p \\ k \end{pmatrix} = \frac{p!}{k!(p-k)!}=\frac{p(p-1)!}{k!(p-k)!}$

It's clear, that $\frac{p!}{k!(p-k)!}=\frac{p(p-1)!}{k!(p-k)!}$ is dividable by p. Thus $p | \frac{p!}{k!(p-k)!}=\frac{p(p-1)!}{k!(p-k)!}$ and thus $ p | \begin{pmatrix}p \\ k \end{pmatrix}$

q.e.d.

Does this work?

ArsenBerk
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xotix
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    you probably mean that $\begin{pmatrix}p \ k\end{pmatrix}=\frac{p!}{k!(p-k)!}$. And yes it does work – Pink Panther Jan 14 '18 at 18:25
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    Any time you're trying to write a proof, if you find yourself writing "It's clear", it's probably not good enough. Also, in this case that's basically the only step in the proof so it's definitely not enough. You should use the definition of divisibility somewhere, for a start. – Sort of Damocles Jan 14 '18 at 18:25
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    Actually you need to justify that $\frac{(p-1)!}{k!(p-k)!}$ is an integer before you can claim this to be a valid proof. – Anurag A Jan 14 '18 at 18:30
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    This doesn't work since it's not a all clear that $\frac{(p-1)!}{k!(p-k)!}$ is an integer. In particular, you're not using the hypothesis that $p$ is prime. If $p$ is not a prime, then this fails: $\frac{1}{4}{4 \choose 2} = \frac{3}{2}$. – MathematicsStudent1122 Jan 14 '18 at 18:30

2 Answers2

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Note that $$ k!\binom{p}{k}=p(p-1)\cdots (p-k+1) $$ Suppose that towards a contradiction, $p$ does not divide $\binom{p}{k}$. Then because $p$ divides the LHS and $p\mid ab\implies p\mid a\quad \text{or}\quad p\mid b$, it follows that $p\mid k!$ so $p\mid j$ for some $1\leq j \leq k<p$, a contradiction.

Sri-Amirthan Theivendran
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Your argument shows that $$\binom{p}{k} =p\left(\frac{(p-1)!}{k!(p-k)!}\right)$$ but it's not immediate that the factor ${\displaystyle{\frac{(p-1)!}{k!(n-k)!}}}$ is an integer.

To finish it, cross-multiply, and then use the primeness of $p$ . . . \begin{align*} \text{Thus,}\;\;&\binom{p}{k} =p\left(\frac{(p-1)!}{k!(n-k)!}\right)\\[4pt] \implies\;&p(p-1)!=(k!)\bigl((p-k)!\bigr)\binom{p}{k}\\[4pt] \end{align*}

Since $p$ is prime,

  • $0 < k < p\implies p\not\mid k!$$\\[4pt]$

    [since $p$ does not divide any of the factors $1,..,k$]$\\[4pt]$

  • $0 < k < p\implies 0 < p - k < p \implies p\not\mid (p-k)!$$\\[4pt]$

    [since $p$ does not divide any of the factors $1,..,p-k$]$\\[4pt]$

In the equation $$p(p-1)!=(k!)((p-k)!)\binom{p}{k}$$ it's clear that $p$ divides the LHS, hence $p$ divides the RHS.

Thus, $p$ divides the product of the $3$ integers $$k!,\;\;\;(p-k)!,\;\;\;\binom{p}{k}$$ hence, since $p$ is prime, $p$ must divide at least one of the $3$ factors.

Since $p\not\mid k!$, and $p\not\mid (p-k)!$, it follows that $p\,{\mid}\,{\large{\binom{p}{k}}}$, as was to be shown.

quasi
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