a) Find all integers $n \in \mathbb{N}^{*}$ that have an odd number of divisors (positive).

b) Let $n \in \mathbb{N}^{*}$. We denote $d$ as the number of divisors of $n$, and $D$ as the product of its divisors. Show that $n^{d}=D^{2}$.
c) Suppose the prime factorization of $n$ is written as $p_{1}^{\alpha_{1}} \ldots p_{r}^{\alpha_{r}}$. Express the number of divisors of $n$ in terms of the $\alpha_{i}$.
a) The key point is to notice that if $d$ is a divisor of $n$, then $\frac{n}{d}$ is also. (In fact, the function $f: d \mapsto \frac{n}{d}$ from the set of divisors of $n$ to itself is an involution, that is, for all $d, f(f(d))=d$). If for every divisor $d$ of $n$, we have $d \neq \frac{n}{d^{\prime}}$, then we can partition the set of divisors of $n$ into pairs $\left\{d, \frac{n}{d}\right\}$, and this set is therefore of even cardinality. Otherwise, there exists a divisor $d_{0}$ of $n$ such that $\frac{n}{d_{0}}=d_{0}$, that is, $n=d_{0}^{2}$, and $n$ is a perfect square. In this case, we can partition the set of divisors of $n$ into a certain number of pairs $\left\{d, \frac{n}{d}\right\}$ and a singleton $\left\{d_{0}\right\}$, and this set is therefore of odd cardinality. The $n \in \mathbb{N}^{*}$ that have an odd number of divisors are therefore exactly the perfect squares.

b) We use the remark from the previous question again. If the set of divisors of $n$ is $\left\{x_{1}, \ldots, x_{d}\right\}$, then we can rewrite this set as $\left\{\frac{n}{x_{1}}, \ldots, \frac{n}{x_{d}}\right\}$. Therefore, $D=x_{1} \cdot \ldots \cdot x_{d}=$ $\frac{n}{x_{1}} \cdot \ldots \cdot \frac{n}{x_{d}}$, from which $D^{2}=\left(x_{1} \frac{n}{x_{1}}\right) \cdot \ldots \cdot\left(x_{d} \frac{n}{x_{d}}\right)=n^{d}$.

c) We notice that a natural number divides $n$ if and only if it can be written as $p_{1}^{\beta_{1}} \ldots p_{r}^{\beta_{r}}$, with $0 \leqslant$ $\beta_{i} \leqslant \alpha_{i}$ for all $i$. The number of divisors of $n$ is therefore equal to the number of possible choices of the sequence of $\beta_{i}$ satisfying these conditions. For a fixed $i$, there are $\alpha_{i}+1$ possible choices for the value of $\beta_{i}$. Therefore, there are $\left(\alpha_{1}+1\right) \ldots\left(\alpha_{r}+1\right)$ possible choices for the entire sequence, and $c^{\prime}$ is the number of divisors of $n$.