For that part, I used Chinese remainder theorem to show that

, 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

be the set of all positive integers less than or equal to

, which are relatively prime to

, and

similar to

with

, then

and

are complete reduced residue systems modulo

and

. Then by the Chinese remainder theorem we have for

, and

, this implies that there exists some

ordered pairs of

such that

and

, hence the above follows.

Does this reasoning make sense?