Is there a positive integer $n$ such that

$$
\left(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{n^{2}-1}+\sqrt{n^{2}}}\right) \cdot\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n^{2}-2}+\sqrt{n^{2}}}\right)=45
$$
From the second factor, it can be seen that $n$ can only be an even number. We simplify each factor in the product on the left side of the equation by rationalizing the denominators in the individual terms. The common denominator in the first factor is 1, and in the second factor, it is 2. We perform the simplifications in the numerators. This results in a quadratic equation for $n$:

\[
\begin{gathered}
\left(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\cdots+\frac{1}{\sqrt{n^{2}-1}+\sqrt{n^{2}}}\right) \cdot\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{4}}+\cdots\right. \\
\left.+\frac{1}{\sqrt{n^{2}-2}+\sqrt{n^{2}}}\right)=\left(\frac{\sqrt{2}-\sqrt{1}}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+\cdots+\frac{\sqrt{n^{2}}-\sqrt{n^{2}-1}}{1}\right) \\
\cdot\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{4}-\sqrt{2}}{2}+\cdots+\frac{\sqrt{n^{2}}-\sqrt{n^{2}-2}}{2}\right)= \\
=\frac{-\sqrt{1}+\sqrt{n^{2}}}{1} \cdot \frac{\sqrt{n^{2}}}{2}=(n-1) \frac{n}{2}=\frac{n^{2}-n}{2}=45
\end{gathered}
\]

From this,

\[
n_{1,2}=\frac{1 \pm \sqrt{1+8 \cdot 45}}{2}
\]

that is,

\[
n_{1}=10 \quad \text { and } \quad n_{2}=-9 \text {. }
\]

The solution $n_{1}=10$ is the only one that satisfies the conditions of the problem, so there is a positive number for which the conditions of the problem are met.

Gábor Perneczky (Kaposvár, Táncsics g. I. o. t.)

Note: The question arises whether the 45 in the problem is a special number, or whether an appropriate $n$ can be found for other values of $A$, and if so, for which values.

For the sought $n$ to be a natural number, the discriminant of the quadratic equation obtained during the solution must be a perfect square, and it must be the square of an odd number, i.e., $A$ must satisfy the following equation:

\[
1+8 A=(2 l+1)^{2}
\]

from which

\[
A=\frac{(2 l+1)^{2}-1}{8}=\frac{4 l^{2}+4 l+1-1}{8}=\frac{l(l+1)}{2}
\]

The condition we derived for $A$ can thus be stated as follows: for every $A$ whose double can be expressed as the product of two consecutive positive integers, there exists a natural number $n$ such that the value of the product in the problem is exactly the given $A$ number.

András Gyene (Bp. VIII., Széchenyi g. II. o. t.)