(solved by Amélie Héliou). If $n$ is an integer, we denote $d(n)$ as the number of positive divisors of $n$. Find all integers $n$ such that $n=d(n)^{2}$.
. Equality directly implies that $n$ is a square, so it can be written in the form $n=p_{1}^{2 \alpha_{1}} p_{2}^{2 \alpha_{2}} \cdots p_{d}^{2 \alpha_{d}}$ where the $\alpha_{i}$ are strictly positive integers and the $p_{i}$ are distinct prime numbers. In this case, we have the usual formula for $d(n)$:

$$
d(n)=\left(2 \alpha_{1}+1\right)\left(2 \alpha_{2}+1\right) \cdots\left(2 \alpha_{d}+1\right)
$$

We first note that this implies that $d(n)$ is odd, and thus by the equality $n=d(n)^{2}, n$ is also odd. In other words, the prime number 2 does not appear among the $p_{i}$, i.e., $p_{i} \geqslant 3$ for all $i$. We have:

$$
1=\frac{d(n)}{\sqrt{n}}=\frac{2 \alpha_{1}+1}{p_{1}^{\alpha_{1}}} \cdot \frac{2 \alpha_{2}+1}{p_{1}^{\alpha_{2}}} \cdots \frac{2 \alpha_{d}+1}{p_{1}^{\alpha_{d}}}
$$

Using $p_{i} \geqslant 3$, it is not difficult to show that each of the preceding factors is greater than 1 and that equality is achieved only if $p_{i}=3$ and $\alpha_{i} \in\{1,2\}$. This leaves only two possibilities for $n$, namely 1 and 9. We verify that both of them are valid.