## Task A-2.2.

Determine all natural numbers $n$ for which there exists a natural number $x$ such that

$$
\frac{1}{\sqrt{x}+\sqrt{x+1}}+\frac{1}{\sqrt{x+1}+\sqrt{x+2}}+\cdots+\frac{1}{\sqrt{x+n}+\sqrt{x+n+1}}=1
$$
## Solution.

By rationalizing, we simplify the given fractions. For $k=0,1, \ldots, n$, we have

$$
\begin{aligned}
\frac{1}{\sqrt{x+k}+\sqrt{x+k+1}} & =\frac{\sqrt{x+k}-\sqrt{x+k+1}}{(\sqrt{x+k}+\sqrt{x+k+1})(\sqrt{x+k}-\sqrt{x+k+1})} \\
& =\frac{\sqrt{x+k}-\sqrt{x+k+1}}{(x+k)-(x+k+1)} \\
& =\sqrt{x+k+1}-\sqrt{x+k}
\end{aligned}
$$

Thus, we obtain the new equality

$$
(\sqrt{x+1}-\sqrt{x})+(\sqrt{x+2}-\sqrt{x+1})+\cdots+(\sqrt{x+n+1}-\sqrt{x+n})=1
$$

or

$$
\sqrt{x+n+1}-\sqrt{x}=1
$$

which means $\sqrt{x+n+1}=\sqrt{x}+1$.

After squaring, we get $x+n+1=x+1+2 \sqrt{x}$, hence $n=2 \sqrt{x}$, or $x=\frac{n^{2}}{4}$.

The number $x$ is a natural number if and only if $n$ is even, i.e., $n=2 k, k \in \mathbb{N}$.