I was looking at review sheet for my midterm for primality testing and factorization and saw this on it
$$4^m < \operatorname{lcm} (m+1, m+2, \dots, 2m+1)$$
How can I prove this?
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Use the facts that $a\cdot b=\gcd(a,b)\cdot \operatorname{lcm}(a,b)$ and that for any $n \in \mathbb{N}$ $\gcd(n,n+1)=1$ – rtybase Nov 12 '16 at 18:50
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My first guess would be to look at some central binomial coefficients $2n \choose n$ – Bruno Joyal Nov 12 '16 at 19:04
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The central binomial coefficient method tends to give upper bounds of the form $4^m$ rather than lower bounds. – Greg Martin Nov 12 '16 at 19:15
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Notice that for $k\leq m$ we have $nk\in\{m+1,\ldots,2m+1\}$ for some $n\geq2$. This imply $$ \text{lcm} \{m+1,\ldots,2m+1\}=\text{lcm}\{m,m+1,\ldots,2m+1\}=\text{lcm}\{m-1,m,m+1,\ldots,2m+1\}=\cdots=\text{lcm}\{1,2,\ldots,2m+1\}. $$
Now we can use that $\text{lcm}\{1,2,\ldots,n\}\geq 2^n$ if $n\geq7$, whose proof you can find in this excellent answer by @shadow10, to conclude $$ \text{lcm}\{m+1,\ldots,2m+1\}\geq 2^{2m+1}=2\cdot4^m>4^m, \hspace{.2cm}m>2. $$ The cases $m=0,1,2$ can be checked by hand.
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1"Notice that for $k \leq m$ we have $2k \in {m+1, \dots, 2m+1}$" - Should $2k$ be a more general $nk$ instead? – Gnuk Nov 12 '16 at 20:45
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Why? Clearly $\text{lcm}{k,m+1,m+2,\ldots,2m+1}\geq\text{lcm}{m+1,m+2,\ldots,2m+1}$. The condition $2k\in{m+1,\ldots,2m+1}$ tells you that this inequality is actually an equality. @HarrySmit – Nov 12 '16 at 20:50
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