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Problem: Let polynomial $Q_n (x) = \frac{x(x-1) \dots (x-n+1)}{n!} \in R[x]$ for some ring $R$. Show that $\forall t \in \mathbb{Z}, Q_n (t) \in \mathbb{Z}$.

My solution: For each $t \in \mathbb{Z}$, we have $t-1, t-2, \dots, t-n+1 \in \mathbb{Z}$, so $\frac{t(t-1) \dots (t-n+1)}{n!} \in \mathbb{Z}$ and $Q_n (t) \in \mathbb{Z}$.

Please check my solution. Thank all!

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marked as duplicate by Servaes, darij grinberg, metamorphy, Yanior Weg, mrtaurho May 22 at 20:39

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  • $\begingroup$ How do you know that each of the factors in the numerator is a multiple of one of the numbers from $1$ through $n$? $\endgroup$ – Aniruddha Deshmukh May 22 at 13:01
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    $\begingroup$ You have just stated that all factors in the numerator are integers. But instead, you should show that the numerator is a multiple of $n!$. To do this, you can read the answers here: The product of $n$ consecutive integers is divisible by $n$ factorial. $\endgroup$ – Minus One-Twelfth May 22 at 13:02
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    $\begingroup$ You proved nothing. The product of $n$ integers needn't be a multiple of $n!$. $\endgroup$ – Yves Daoust May 22 at 13:02
  • $\begingroup$ The combinatorial argument: $Q_n(x)=\binom xn$ is the number of ways of choosing $n$ out of $x\ge n$ objects. For $0\le x<n$, one of the factors of the numerator is $0$. For $x<0$, the argument is similar to the one for $x\ge n$. $\endgroup$ – Shubham Johri May 22 at 13:03
  • $\begingroup$ I knew this formulate $\binom{m}{n} = \frac{n(n-1)\dots(n-m+1)}{m!}$. But with this formulate, the prove is clear. $\endgroup$ – Minh May 22 at 13:09
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As shown here:
Binomial coefficient generalization
where $(\cdot )_{k}$ is the Pochhammer symbol, here standing for a falling factorial. So,
$$Q_n (x) = \frac{x(x-1) \dots (x-n+1)}{n!} = \frac {(x)_n} {n!} = \frac {\frac {x!} {(x-n)!}} {n!} = \frac {x!} {n! (x-n)!} = \binom x n \in R[x]$$ Now we know that $x$ is an integer so: $\forall t \in \mathbb{Z}, Q_n (t) \in \mathbb{Z}$.

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    $\begingroup$ I know this formula but I cannot prove it $\endgroup$ – Minh May 22 at 15:41

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