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Here's a problem that I heard: Show that $lcm(1,2,\ldots,2n+1)\geq4^n$.

If you assume the prime number theorem then it's not too hard to show that $lcm(1,2,\ldots,2n+1)$ is eventually larger than $4^n$. The hard part is making this effective for small values of $n$. Also, I've been told that you can prove this without the prime number theorem.

Thomas Browning
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Here's a cool way to do this. Consider the following integral $$\text{lcm}(1,2,3,\cdots,2n+1)\int_0^1x^n(1-x)^ndx.$$

We can expand the product $x^n(1-x)^n$ using the binomial theorem as $$x^n\sum_{k=0}^n \binom{n}{k}(-1)^kx^k = \sum_{k=0}^n \binom{n}{k}(-1)^kx^{n+k}$$

which gives the integral as $$\text{lcm}(1,2,3,\cdots,2n+1) \left(\sum_{k=0}^{n} \binom{n}{k}(-1)^k \frac{x^{n+k+1}}{n+k+1}\right)\Bigg{|}_{0}^1. $$

Now this expression is an integer since the term $\text{lcm}(1,2,3,\cdots,2n+1)$ will clear all the denominators of the sum. This integer is also clearly positive since the function $x^n(1-x)^n$ is positive on $(0,1)$. So we conclude that $$\text{lcm}(1,2,3,\cdots,2n+1)\int_0^1x^n(1-x)^ndx \geq 1.$$

Now using simple calculus, we see that the maximum of the function $x^n(1-x)^n$ on $[0,1]$ occurs at $x = \frac{1}{2}$, and so $x^n(1-x)^n <\frac{1}{4^n}$. This immediately gives that

$$1 \leq \text{lcm}(1,2,3,\cdots,2n+1)\int_0^1x^n(1-x)^ndx < \text{lcm}(1,2,3,\cdots,2n+1)\frac{1}{4^n}$$

and so $$4^n < \text{lcm}(1,2,3,\cdots,2n+1)$$

user413766
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    Very elegant solution! – Thomas Browning Dec 07 '19 at 05:28
  • Thank you! Though I must confess that I didn't come up with this on my own, we did something similar in one of my analytic number theory course at BYU. – user413766 Dec 07 '19 at 05:30
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    For reference, here's the simple calculus: the derivative of $x^n(1-x)^n$ is $nx^{n-1}(1-x)^{n-1}(1-2x)$ which only vanishes at ${0,1/2,1}$. The function $x^n(1-x)^n$ vanishes at $0$ and $1$ so the maximum must be at $1/2$. – Thomas Browning Dec 07 '19 at 05:31
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    One can also see without calculus that $x=1/2$ maximizes $x^n(1-x)^n$ on $[0,1]$ since it maximizes $x(1-x)$. – arkeet Dec 07 '19 at 05:37
  • Hmm. I gave a bounty to this pretty solution yesterday. Probably accidental duplication, but a bit careless nevertheless. – Jyrki Lahtonen Dec 07 '19 at 06:44