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I started by induction on $k$

For $k=1$ then : $1\in \mathbb{N}$

For $k=2$ then : $\frac{(p-1)}{2!} \in \mathbb{N}$ , indeed for all $p>{2}$, $p-1$ is even. (We still have $k<p$ it's important).

Suppose that $\frac{(p-1)...(p-k+1)}{k!} \in \mathbb{N}$ for a specific $k \in \{1,..,p-1\}$

Now try to prove that : $\frac{(p-1)...(p-k)}{(k+1)!} \in \mathbb{N}$.

By the recursive formula we have directly :

$\frac{\binom{p}{k+1}}{p} = \frac{[\binom{p+1}{k+1} - \binom{p}{k}]}{p}$ $\Leftrightarrow$ $\frac{(p-1)...(p-k)}{(k+1)!} = \frac{(p+1)(p-1)...(p-k+1)}{(k+1)!} - \frac{(p-1)...(p-k+1)}{k!}$

By hypothesis we know that $\frac{(p-1)...(p-k+1)}{k!} \in \mathbb{N}$

Now, the key is to see that a product of $k$ consecutive terms is divided by $k!$ (a little induction can solve this).

Then the induction is complete.

Maman
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  • Hi. In the first line, when you use Pascal recursion, you should use $-$ instead of "+", does not it? If you do that, you will get $\frac{(p+1)-(k+1)}{k+1}=\frac{p-k}{k+1}$. Do not know if that is the right track. – Phicar Dec 03 '14 at 23:59
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    Do you need to use induction? Otherwise there may be a direct proof that $\frac{{p \choose k}}{p}\in \mathbb{N}$ – Kaladin Dec 04 '14 at 00:00
  • @Phicar No you're right it is "-" but I still have the same problem for $k=1$. – Maman Dec 04 '14 at 00:08
  • @Kaladin There is a solution which consists to apply Gauss Lemma but I want to prove the result by induction. – Maman Dec 04 '14 at 00:09
  • You will have a lot of problems, not just for $1$. I suggest you to look further in $\frac{(p+1)(p-1)\ldots (p-k+1)}{(k+1)!}$.. you take out $p$, but $k+1<p$, so who will divide $p$? – Phicar Dec 04 '14 at 00:14
  • I don't think $\frac{p-k}{k+1}\in\mathbb{N}$ take $k>\lfloor p/2\rfloor$ then $\frac{p-k}{k+1}<1$ – Kaladin Dec 04 '14 at 00:22
  • @Phicar $p$ is a prime number so $1$ and $p$ can divide $p$ – Maman Dec 04 '14 at 00:29
  • Do you need to use induction? Because if $\binom{p}{k}$ is an integer and $p$ divides $\binom{p}{k}$... – Aaron Maroja Dec 04 '14 at 00:45
  • @AaronMaroja For me to prove that $p$ divides $\binom{p}{k}$ is not clear – Maman Dec 04 '14 at 00:47
  • @Maman Well, $p$ is prime, and $\binom{p}{k}$ is an integer, suppose a factor of $k!$ divides $p$ , where is the contradiction? – Aaron Maroja Dec 04 '14 at 00:53
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    Did you take a look here? – Aaron Maroja Dec 04 '14 at 00:56
  • @AaronMaroja the fact is that $p$ is prime with every $k \in {1,..,p-1}$ – Maman Dec 04 '14 at 00:57
  • That's right, then $p$ divides $\binom{p}{k}$. – Aaron Maroja Dec 04 '14 at 00:59
  • @AaronMaroja I already knew this method but I just wanted to prove it by induction just to change the point of view (more formal) – Maman Dec 04 '14 at 00:59
  • I see, you could use valuation theory to do so, see the answer on the question I showed, really interesting approach. – Aaron Maroja Dec 04 '14 at 01:00
  • @AaronMaroja maybe with my try we can prove that $\frac{(p+1)(p-1)\ldots (p-k+1)}{(k+1)!} \in \mathbb{N}$ but I don't see the right argument. – Maman Dec 04 '14 at 16:53
  • @AaronMaroja Is that true the fact $k!$ divide the product of $k$ numbers consecutively ? – Maman Aug 30 '16 at 16:24

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By definition, $$\binom{p}{k} = \frac{p!}{k!(p-k)!}$$ Since $p$ is prime, it's relative prime to all smaller numbers, and therefore for $k<p$, it cannot be a factor of $k!$. Now if further $k>0$, then $p-k<p$, and thus $p$ is not a factor of $k!(n-k)!$. Since by definition, $p$ is a factor of $p!$, and $p$ is prime, it follows that it is a factor of $\binom{p}{k}$, thus $\binom{p}{k}/p\in\mathbb N$.

celtschk
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  • I think here should be reliance on to the fact $\binom{p}{k} \in \mathbb{N}$. However, +1. – Tacet Dec 06 '14 at 14:30
  • It's right but as I said in the comments before I want to prove that $\frac{(p+1)(p-1)\ldots (p-k+1)}{(k+1)!}$ is an integer to complete the induction. – Maman Dec 06 '14 at 14:52