If $m$ and $n$ are integers with $\gcd(m,n) = 1$, prove that $\sigma(mn)= \sigma(m)\sigma(n)$. I am thinking about using the formula for $\sigma(p^k)$ where $p$ is prime.
It follows from the geometric series that
$$\sigma(p^k) = 1 + p + p^2 +\cdots+p^k = \frac{p^{k+1}-1}{p-1}$$
Any Help?
Let $D(a)$ is the set of all positive divisors of $a$.
When $\gcd(m,n)=1$, the map $(x,y) \mapsto xy$ defines a bijection $D(m)\times D(n)\to D(mn)$.
Use this bijection when you expand $\sigma(m)\sigma(n)=\big(\sum_{x\in D(m)} x\big)\cdot \big(\sum_{y\in D(n)} y\big)$ to conclude that it is the same as $\sigma(mn)= \sum_{z\in D(mn)}z$.
Let $m = p_1^{e_1}p_2^{e_2}...p_k^{e_k}$ and $n = q_1^{d_1}q_2^{d_2}...q_i^{d_i}$ for $p, q$ prime and $e, d \in \mathbb{Z}$.
Then we have $$\sigma(mn) = \sigma(p_1^{e_1}p_2^{e_2}...p_k^{e_k}*q_1^{d_1}q_2^{d_2}...q_i^{d_i})$$
We know the divisors of $p_1^{e_1}$ are $1, p_1, p_1^2, p_1^3,..., p_1^{e_1}.$ This holds for any prime to any power.
Since the gcd$(m,n) = 1$, we also know $m$ and $n$ share no common divisors
So we have $$=[1+p_1+p_1^2+...+p_1^{e_1}][1+p_2+p_2^2+...+p_2^{e_2}]...[1+p_k+p_k^2+...+p_k^{e_k}] * [1+q_1+q_1^2+...+q_1^{d_1}][1+q_2+q_2^2+...+q_2^{d_2}]...[1+q_i+q_i^2+...+q_i^{d_i}]$$ The first half of the statement above is just the divisors of $m$ and the second half is the divisors of $n$
Then $$= \sigma(m)\sigma(n) $$
The proof is bidirectional
$(m,n)=1$, $\sigma(mn)=\sum_{d|mn} d$ then if d|mn $\rightarrow$ $d=ab$ with $a|m $ , $b|n $, $(a,b)=1$ then $\sigma(mn)=\sum_{ab|mn}ab=\sum_{a|m}a\sum_{b|n}b=\sigma(m)\sigma(n)$.
