nicetry007 wrote: The product $ (c_{m+1}-c_{k})\; (c_{m+2}-c_{k})\;\cdots\; (c_{m+n}-c_{k})\;$ $ = \left((m+1)(m+2)-k(k+1)\right)\;\left((m+2)(m+3)-k(k+1)\right)\;\cdots \left((m+n)(m+n+1)-k(k+1)\right)$ $ = (m+1-k)(m+k+2)\; (m+2-k)(m+k+3)\;\cdots\;(m+n-k)(m+k+n+1)$ $ = \Pi_{i =1}^{n}(m+i-k) \;\Pi_{i=1}^{n}(m+k+i+1)$ The product $ c_{1}\;c_{2}\; \cdots \; c_{n}= (1 \cdot 2)\; (2 \cdot 3)\; \cdots \; (n \cdot (n+1))\; = n! (n+1)!$ The ratio $ \frac{(c_{m+1}-c_{k})\; (c_{m+2}-c_{k})\;\cdots\; (c_{m+n}-c_{k})\;}{c_{1}\;c_{2}\; \cdots \; c_{n}}$ $ = \frac{\Pi_{i =1}^{n}(m+i-k) \;\Pi_{i=1}^{n}(m+k+i+1)}{n! (n+1)!}$ $ = \;^{(m+n-k)}C_{n}\;\cdot \; \left( \frac{1}{m+k+1}\right) \;\cdot \;^{(m+k+n+1)}C_{ n+1}$ Since $ n+1\; < \;m+k+1$, the binomial term $ ^{(m+k+n+1)}C_{ n+1}$ is divisible by $ (m+k+1)$. Hence, $ \left( \frac{1}{m+k+1}\right) \;^{(m+k+n+1)}C_{ n+1}$ is an integer and the product $ (c_{m+1}-c_{k})\; (c_{m+2}-c_{k})\;\cdots\; (c_{m+n}-c_{k})\;$ is divisible by $ c_{1}\;c_{2}\; \cdots \; c_{n}$.

Could you explain me how you reduced $ \frac{\Pi_{i =1}^{n}(m+i-k) \;\Pi_{i=1}^{n}(m+k+i+1)}{n! (n+1)!}$ to: $ \;^{(m+n-k)}C_{n}\;\cdot \; \left( \frac{1}{m+k+1}\right) \;\cdot \;^{(m+k+n+1)}C_{ n+1}$ Thank you.

Yeah, that reduction is wrong indeed. I also think an extra condition $ m>k$ or $ m+n<k$ wouldould be necessary, otherwise we have to show $ 0|c_1\cdots c_n$, which is obviously false.

Write $ m + k + 1 =: p$ (a prime) $ > n + 1$. Note that $ c_{1}c_{2} \cdots c_{n} = (1 \cdot 2) (2 \cdot 3) \cdots (n \cdot (n + 1)) = n! (n + 1)! = (n + 1)(n!)^2$ whilst $ \begin{array}{cl} & (c_{m + 1} - c_{k}) (c_{m + 2} - c_{k})\cdots (c_{m + n} - c_{k})\\ \mbox{ =} & \left((m + 1)(m + 2) - k(k + 1)\right)\left((m + 2)(m + 3) - k(k + 1)\right)\cdots \left((m + n)(m + n + 1) - k(k + 1)\right)\\ \mbox{ =} & (2\Sigma_{j=k+1}^{m+1}j) (2\Sigma_{j=k+1}^{m+2}j) \cdots (2\Sigma_{j=k+1}^{m+n}j)\\ \mbox{ =} & (m + 1 - k)(p + 1) (m + 2 - k)(p + 2)\cdots(m + n - k)(p + n)\\ \mbox{ =} & \Pi_{i = 1}^{n}(m + i - k) \Pi_{i = 1}^{n}(p + i)\\ \mbox{ =} & {{m-k+n} \choose {m-k}}n!{{p+n} \choose p}n!\\ \mbox{ =} & {{m-k+n} \choose {m-k}}{{p+n} \choose p}(n!)^2\\ \end{array}$ Now I am going to show that $ (n+1) \left| {{p+n} \choose p} \right.$ i.e. $ \frac{1}{n+1} {{p+n} \choose p} \in \mathbb{Z}$ and we are done. $ \frac{1}{n+1} {{p+n} \choose p} = \frac{(p+n)!}{(n+1)!p!} = \frac{1}{p}\frac{(p+n)!}{(n+1)!(p-1)!} = \frac{1}{p}{{p+n}\choose{p-1}}$ Note that $ p | (p+n)!$ (since $ p<p+n$), but $ p \not| (n+1)!$ (since $ n+1<p$) and $ p \not| (p-1)!$ (since $ p-1<p$). So the prime $ p \left|\frac{(p+n)!}{(n+1)!(p-1)!} \right.$ $ \blacklozenge$

Phill and Peter, The reduction is not wrong. If $ m+1 \leq k \leq m+n$, then the result is trivially true as $ c_1c_2...c_n|0$. If $ k < m+1$, then follow the proof from the second post.

Quote: $ = (m+1-k)(m+k+2)\; (m+2-k)(m+k+3)\;\cdots\;(m+n-k)(m+k+n+1) $ or Quote: $ = (m+1-k)(p+1) (m+2-k)(p+2)\cdots(m+n-k)(p+n)$ This I can't see how can be obtained. Could anybody explain the details?