I have a question given to me by a friend.
Prove that $\frac {(m,n){n\choose m}}{n}$ is an integer for all integers $n\ge m\ge1$. Any hint??
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Use bezout to express $(m,n)$ as $am+bn$ – Vidyanshu Mishra Sep 25 '17 at 13:21
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What have you tried? Also, is it $$\frac{(m, n)\binom{n}{m}}{m}$$ or is it $$\frac{(m, n)\binom{n}{m}}{n}?$$ – Shaun Sep 25 '17 at 13:21
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Sorry, I m edit it. – Narayan Sep 25 '17 at 13:23
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One is divided by $m$; the other, $n$. – Shaun Sep 25 '17 at 13:23
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1@Shaun, now good? – Narayan Sep 25 '17 at 13:24
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By $(m,n)$ do you mean gcd$(m,n)$? – Teddy38 Sep 25 '17 at 13:24
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Yes, @Narayan ${}$ – Shaun Sep 25 '17 at 13:24
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Yes @Edmo38.... – Narayan Sep 25 '17 at 13:25
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I was trying to remember where I have seen this problem. This is Putnam $2000$ problem – Vidyanshu Mishra Sep 25 '17 at 13:37
1 Answers
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By Bezout’s identity, there exist integers $a$ and $b$ such $gcd(m, n) = am + bn$.
Next, notice that $$\frac{gcd(m, n)\times {n\choose m}}{n}=\frac{(am+bn)\times {n\choose m}}{n}=\frac{am\times {n\choose m}}{n}+b\times {n\choose m}= a\times {n−1 \choose m−1}+b\times{n\choose m} $$
Which is clearly an integer as $n\ge m\ge1$.
Vidyanshu Mishra
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