1. ## Divisible integers

Here's the question:
Find all positive integers n divisible by all positive integers m where m is less than or equal to n^(1/2)

2. First note that the only $n<36$ which fill the condition are

$n = 1, 2, 3, 4, 6, 8, 12, 24$. We'll see there are no others.

Suppose $n$ fills the conditions, and let $t = n-\lfloor \sqrt n \rfloor^2$.

We can suppose now that $\lfloor \sqrt n \rfloor \ge 6$ and then arrive at a contradiction.

$\lfloor \sqrt n \rfloor \mid n$, so there exists a positive integer $s$ such that

$s\lfloor \sqrt n \rfloor = n = \lfloor \sqrt n \rfloor^2+t$

Therefore $\lfloor \sqrt n \rfloor\mid t$ and we can write $n-\lfloor \sqrt n \rfloor^2 = t = d\lfloor \sqrt n \rfloor$

That is, $n = \lfloor \sqrt n \rfloor^2+d\lfloor \sqrt n \rfloor$.

Write $\lfloor \sqrt n \rfloor = \sqrt{n}-\epsilon,\:\: 0\le\epsilon<1$. Plugging in the previous equality yields:

$d = \epsilon\frac{2\sqrt{n}-\epsilon}{\sqrt{n}-\epsilon} < \frac{2\sqrt{n}-\epsilon}{\sqrt{n}-\epsilon} = 1+\frac{\sqrt{n}}{\sqrt{n}-\epsilon} < 1+ \frac{\sqrt{n}}{\sqrt{n}-2}$

And since $\sqrt{n}\ge 6$ we have $\frac{\sqrt{n}}{\sqrt{n}-2}<2$, so $d<1+2 = 3 \Rightarrow d\in\{0,1,2\}$

Thus $n\le h^2+2h$ where we denote $h=\lfloor \sqrt n \rfloor$

We have $h\mid n$ and $h-1\mid n$ and these numbers

are coprime so $h(h-1)\mid n$. Furthermore, $h-2\mid n$

and is coprime to $h-1$, and $\gcd(h-2,h)\le2$,

so $h(h-1)(h-2) \mid 2n$, and thus $h(h-1)(h-2) \le 2n \le 2h^2+4h$

Therefore $h^2-3h+2 \le 2h+4 \Rightarrow h^2-5h -2 \le 0 \Rightarrow h < 6$,

contradicting our assumption that $h=\lfloor\sqrt{n}\rfloor \ge 6$ (see head of the post).