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**gaussrelatz** For that part, I used Chinese remainder theorem to show that $\displaystyle \sigma (m_1 m_2)=\sigma(m_1) \sigma(m_2)$, where the proof I wrote goes like:

proof:

$$\sigma (m_1 m_2)=\sum_{a=1,(a,m_1m_2)=1}^{m_1m_2}(a-1,m_1m_2)$$

$$\sigma (m_1 m_2)=\sum_{a=1,(a,m_1m_2)=1}^{m_1m_2}(a-1,m_1)(a-1,m_2)$$

$$\sigma (m_1 m_2)=\sum_{a \in A, b \in B}(\alpha-1,m_1)(\beta-1,m_2)$$

$$\sigma (m_1 m_2)=\sum_{a \in A}(\alpha-1,m_1)\sum_{ b \in B}(\beta-1,m_2)$$

$$\sigma (m_1 m_2)=\sigma (m_1)\sigma (m_2)$$

Further remarks, if we let $\displaystyle A $ be the set of all positive integers less than or equal to $\displaystyle m_1$ , which are relatively prime to $\displaystyle m_1 $, and $\displaystyle B $ similar to $\displaystyle A$ with $\displaystyle m_2$, then $\displaystyle A$ and $\displaystyle B$ are complete reduced residue systems modulo $\displaystyle m_1$ and$\displaystyle m_2$. Then by the Chinese remainder theorem we have for $\displaystyle 1 \le a \le m_1m_2$, and $\displaystyle (a, m_1m_2)=1$, this implies that there exists some $\displaystyle !$ ordered pairs of $\displaystyle (a,b) \in A \times B$ such that $\displaystyle a = \alpha \bmod m_1$ and $\displaystyle a=\beta \bmod m_2 $, hence the above follows.

Does this reasoning make sense?