11.7. Initially, a natural number $N$ is written on the board. At any moment, Misha can choose a number $a>1$ on the board, erase it, and write down all natural divisors of $a$, except for $a$ itself (the same numbers can appear on the board). After some time, it turned out that there are $N^{2}$ numbers on the board. For which $N$ could this have happened?

(M. Fadin, K. Kovalenko)
Answer. Only for $N=1$.

Solution. Lemma. For any natural $n>1$, the inequality holds

$$
\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{n^{2}}<1
$$

Proof. Consider the function $f(t) = \frac{1}{t^{2}}$ for $t > 1$. The function $f(t)$ is decreasing for $t > 1$. Let $1 = d_{1} < d_{2} < \ldots < d_{k} < d_{k+1} = N$ be the divisors of the number $N$. Then the numbers $N / d_{1} > N / d_{2} > \ldots > N / d_{k} > N / d_{k+1}$ are also divisors of the number $N$. Therefore, they are equal to the numbers $d_{k+1}, d_{k}, \ldots, d_{2}, d_{1}$ respectively. Taking the lemma into account, we get that the total number of numbers on the board will not exceed the number

$$
\begin{aligned}
d_{1}^{2}+d_{2}^{2} & +\ldots+d_{k}^{2}=\frac{N^{2}}{d_{k+1}^{2}}+\frac{N^{2}}{d_{k}^{2}}+\ldots+\frac{N^{2}}{d_{2}^{2}}= \\
& =N^{2}\left(\frac{1}{d_{2}^{2}}+\frac{1}{d_{3}^{2}}+\ldots+\frac{1}{d_{k+1}^{2}}\right) \leqslant \\
& \leqslant N^{2}\left(\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{N^{2}}\right)<N^{2} \cdot 1=N^{2}
\end{aligned}
$$

which is what we needed to prove.