Let $n$ be an integer greater than or equal to $1$. Find, as a function of $n$, the smallest integer $k\ge 2$ such that, among any $k$ real numbers, there are necessarily two of which the difference, in absolute value, is either strictly less than $1 / n$, either strictly greater than $n$.
$\textsf{Claim: } k \geq n^2+2$
$\textsf{Proof: }$ Let $d$ be the minimum value of the difference between $2$ terms of the sequence of $k$ nos.
First see that, if $k < n^2+2$, consider the $k$ numbers $0,\frac{1}{n},\frac{2}{n},...,\frac{k-1}{n} \in \mathbb{R}$. Here, maximum difference of two terms is $=\frac{k-1}{n}$ and minimum is $=\frac{1}{n}$. So, the bounds are not strict.
Now, consider $k=n^2+2$.
Let the terms in the sorted sequence be $a_0,a_1,...,a_{k-1} \in \mathbb{R}$.
Now, observe that $a_p \geq a_0 + dp, \forall p <k$
By Induction, Base case is trivial.
Let the induction hypothesis be $a_{k-2} \geq a_0+(k-2)d$
Now, we need to prove that $a_{k-1} \geq a_0+(k-1)d$
So, we can say that, $a_{k-1} \geq a_{k-2}+d =a_0+(k-1)d$ (proved, since $d$ is the minimum value of the difference)
So, $a_{k-1} \geq a_0+(k-1)d \implies a_{k-1}-a_0 \geq (k-1)d=(n^2+1)d > n^2d >n$ or $d <\frac{1}{n}$. (can be checked by putting $d=\frac{1}{n}$)