If $n \in \mathbb{N}^*$, we denote, for $i \in \{0,1,2,3\}$, $$r_i(n) = \operatorname{Card}\{d \in \mathbb{N} : d \equiv i [4] \text{ and } d \mid n\}$$ We set $g(n) = r_1(n) - r_3(n)$.
Show that if $m$ and $n$ are two nonzero natural integers that are coprime, we have $g(mn) = g(m)g(n)$.

Let $d_1$ and $d_2$ be two coprime integers. Show that for all $(u,v)\in\mathbb{Z}^2$, $$L_{\mathrm{prim}}(u,v,d_1 d_2) = L_{\mathrm{prim}}(u,v,d_1)\,L_{\mathrm{prim}}(u,v,d_2).$$

Let $p$ be a prime number and $\alpha\geq 1$ be an integer. Show that $$L_{\mathrm{prim}}(u,v,p^\alpha) = L(u,v,p^\alpha) - L(u,v,p^{\alpha-1}).$$

For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ Prove, using a summation interchange, that $\frac{1}{x} \sum_{n \leqslant x} \omega(n) \underset{x \rightarrow +\infty}{=} \ln_{2}(x) + O(1)$.

For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ Show that $$\frac{1}{x} \sum_{n \leqslant x} \left(\omega(n) - \ln_{2}(x)\right)^{2} \underset{x \rightarrow +\infty}{=} \frac{1}{x}\left(\sum_{n \leqslant x} \omega(n)^{2}\right) - \ln_{2}(x)^{2} + O\left(\ln_{2}(x)\right).$$

For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ Show that $$\sum_{n \leqslant x} \omega(n)^{2} = \sum_{\substack{p_{1} \leqslant x \\ p_{1} \text{ prime}}} \sum_{\substack{p_{2} \leqslant x \\ p_{2} \text{ prime}}} \operatorname{Card}\left\{n \in \mathbb{N}^{*} : n \leqslant x, p_{1} \mid n \text{ and } p_{2} \mid n\right\}$$

For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ Conclude that $\frac{1}{x}\left(\sum_{n \leqslant x} \left(\omega(n) - \ln_{2}(x)\right)^{2}\right) \underset{x \rightarrow +\infty}{=} O\left(\ln_{2}(x)\right)$.

For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ Prove, using a change of order of summation, that $\frac{1}{x} \sum_{n \leqslant x} \omega(n) \underset{x \rightarrow +\infty}{=} \ln_2(x) + O(1)$.

For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ Show that $$\frac{1}{x} \sum_{n \leqslant x} \left(\omega(n) - \ln_2(x)\right)^2 \underset{x \rightarrow +\infty}{=} \frac{1}{x}\left(\sum_{n \leqslant x} \omega(n)^2\right) - \ln_2(x)^2 + O\left(\ln_2(x)\right)$$

For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ Show that $$\sum_{n \leqslant x} \omega(n)^2 = \sum_{\substack{p_1 \leqslant x \\ p_1 \text{ prime}}} \sum_{\substack{p_2 \leqslant x \\ p_2 \text{ prime}}} \operatorname{Card}\left\{n \in \mathbb{N}^* : n \leqslant x, p_1 \mid n \text{ and } p_2 \mid n\right\}$$

For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ Conclude that $\frac{1}{x}\left(\sum_{n \leqslant x} \left(\omega(n) - \ln_2(x)\right)^2\right) \underset{x \rightarrow +\infty}{=} O\left(\ln_2(x)\right)$.

Let $g(n) = 5^k$ where $k$ is the number of distinct primes dividing $n$, and let $h(n) = 0$ if $n$ is divisible by $k^2$ for some integer $k > 1$, and $h(n) = 1$ otherwise.
a) Show that $g(mn) = g(m)g(n)$ does not hold in general, and determine when it holds.
b) Show that $h(mn) = h(m)h(n)$ for all positive integers $m, n$.

Let $d_1, d_2, \ldots, d_k$ be all the factors of a positive integer $n$ including 1 and $n$. If $d_1 + d_2 + \ldots + d_k = 72$, then $\frac{1}{d_1} + \frac{1}{d_2} + \cdots + \frac{1}{d_k}$ is:
(A) $\frac{k^2}{72}$
(B) $\frac{72}{k}$
(C) $\frac{72}{n}$
(D) none of the above

Let $d _ { 1 } , d _ { 2 } , \ldots , d _ { k }$ be all the factors of a positive integer $n$ including 1 and $n$. If $d _ { 1 } + d _ { 2 } + \ldots + d _ { k } = 72$, then $\frac { 1 } { d _ { 1 } } + \frac { 1 } { d _ { 2 } } + \cdots + \frac { 1 } { d _ { k } }$ is:
(a) $\frac { k ^ { 2 } } { 72 }$
(b) $\frac { 72 } { k }$
(c) $\frac { 72 } { n }$
(d) none of the above.

Let $d _ { 1 } , d _ { 2 } , \ldots , d _ { k }$ be all the factors of a positive integer $n$ including 1 and $n$. If $d _ { 1 } + d _ { 2 } + \ldots + d _ { k } = 72$, then $\frac { 1 } { d _ { 1 } } + \frac { 1 } { d _ { 2 } } + \cdots + \frac { 1 } { d _ { k } }$ is:
(a) $\frac { k ^ { 2 } } { 72 }$
(b) $\frac { 72 } { k }$
(c) $\frac { 72 } { n }$
(d) none of the above.

Let $d_1, d_2, \ldots, d_k$ be all the factors of a positive integer $n$ including 1 and $n$. If $d_1 + d_2 + \ldots + d_k = 72$, then $\frac{1}{d_1} + \frac{1}{d_2} + \cdots + \frac{1}{d_k}$ is:
(A) $\frac{k^2}{72}$
(B) $\frac{72}{k}$
(C) $\frac{72}{n}$
(D) none of the above

Let $d _ { 1 } , d _ { 2 } , \ldots , d _ { k }$ be all the factors of a positive integer $n$ including 1 and $n$. If $d _ { 1 } + d _ { 2 } + \ldots + d _ { k } = 72$, then $\frac { 1 } { d _ { 1 } } + \frac { 1 } { d _ { 2 } } + \cdots + \frac { 1 } { d _ { k } }$ is:
(A) $\frac { k ^ { 2 } } { 72 }$
(B) $\frac { 72 } { k }$
(C) $\frac { 72 } { n }$
(D) none of the above

For any positive integer $n$, and $i = 1, 2$, let $f_i(n)$ denote the number of divisors of $n$ of the form $3k + i$ (including 1 and $n$). Define, for any positive integer $n$, $$f(n) = f_1(n) - f_2(n)$$ Find the values of $f\left(5^{2022}\right)$ and $f\left(21^{2022}\right)$.

Let $S ^ { 1 } = \{ z \in \mathbb { C } | | z \mid = 1 \}$ be the unit circle in the complex plane. Let $f : S ^ { 1 } \rightarrow S ^ { 1 }$ be the map given by $f ( z ) = z ^ { 2 }$. We define $f ^ { ( 1 ) } : = f$ and $f ^ { ( k + 1 ) } : = f \circ f ^ { ( k ) }$ for $k \geq 1$. The smallest positive integer $n$ such that $f ^ { ( n ) } ( z ) = z$ is called the period of $z$. Determine the total number of points in $S ^ { 1 }$ of period 2025. (Hint: $2025 = 3 ^ { 4 } \times 5 ^ { 2 }$)