For positive integers $n$, let $x_n, y_n$ be defined as:
$x_n = \sqrt{n} + \sqrt{n+1}$ and $y_n = \sqrt{4n + 2}$
Show that there is no whole number between $x_n$ and $y_n$.
Observation: Obviously $x_n \not\in \mathbb{N}$ since $n$ and $n+1$ cannot both be perfect squares and $y_n \not\in \mathbb{N}$ since $4n +2 \equiv 2 \textrm{ (mod 4)}$ and any square must be $\equiv 0 \textrm{ or } 1 \textrm{ (mod 4)}$.
Question from 2009 German Math Olympiad, 2nd round
A very straightforward approach can be as below;
Assume $m^2\lt n \le (m+1)^2$. Hence;
$$2m \lt \sqrt n+ \sqrt {n+1}\lt \sqrt {4n+2} \lt 2m+3.$$
Therefore $k$ (the integer between $x_n$ and $y_n$) is either $2m+1$ or $2m+2$. But It is impossible for $k$ to be $2m+2$ because, in this manner, $n$ has to be $(m+1)^2$ (note that $\sqrt {4((m+1)^2-1)+2} \lt 2m+2$) while;
$$(m+1)+(m+1)\lt \sqrt n+ \sqrt {n+1}\lt 2m+2,$$ which is a contradiction.
Now, if $k=2m+1$, we must have:
$$\sqrt {4n+2}\gt 2m+1\implies n\gt m^2+m-\frac{1}{4} \implies n\ge m^2+m. $$
But, in this case, we must also have:
$$\sqrt{m^2+m}+\sqrt {m^2+m+1}\le \sqrt n+ \sqrt {n+1}\lt 2m+1, $$
which is impossible; because for every integer $m$, we have:
$$\sqrt{m^2+m}+\sqrt {m^2+m+1}\gt 2m+1.$$
