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I am trying to prove that $\sigma(p_1^a\cdot p_2^b) =\sigma(p_1^a)\cdot\sigma(p_2^b)$ where $p_1$ and $p_2$ are prime numbers.
We know that $\sigma(p_1^a) = \frac{p_1^{a+1}-1}{p_1-1}$ and $\sigma(p_2^b) = \frac{p_2^{b+1}-1}{p_2-1}$.
Now I am trying to find the divisors of $p_1^a\cdot p_2^b$ and add them: I found that the divisors are

$1$, $p_1$, $p_1^{2},\dotsc,p_1^{a}$, $p_2$, $p_2^{2},\dotsc,p_2^{b}$, $p_1\cdot p_2$, $p_1\cdot p_2^2,\dotsc,p_1\cdot p_2^{b},\dotsc,p_1^{a}\cdot p_2,\dotsc,p_1^{a}\cdot p_2^{b}$.

Now when we do their summation we get $\sum_{k=0}^a$ $p_1^k$ + $\sum_{k=1}^b$ $p_2^k$ + ($\sum_{k=1}^{a}$ $p_1^k\cdot\sum_{k=1}^b$ $p_2^k$), is this right? If yes I can't reach $\frac{p_1^{a+1}-1}{p_1-1}\cdot\frac{p_2^{b+1}-1}{p_2-1}$.

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You are very close. Just notice that $$\sum_{k=0}^a p_1^k + \sum_{k=1}^b p_2^k + \left(\sum_{k=1}^{a} p_1^k\cdot\sum_{k=1}^b p_2^k \right) $$ equals $$ 1+\sum_{k=1}^a p_1^k + \sum_{k=1}^b p_2^k + \left(\sum_{k=1}^{a} p_1^k\cdot\sum_{k=1}^b p_2^k \right)=\sum_{k=0}^a p_1^k \cdot \sum_{k=0}^b p_2^k $$ which is $$ \sigma(p_1^a) \sigma(p_2^b) $$ so you are done.

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  • $\begingroup$ Yes, i figured it out. Thank you $\endgroup$ – outlawoutlawz Oct 1 '16 at 5:56
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In general if the prime factorization of $n$ is $$n=p_1^{a_1}p_2^{a_2}\dotsm p_m^{a_m},$$ then the divisors of $n$ will be of the form $$p_1^{b_1}p_2^{b_2}\dotsm p_m^{b_m}\quad (0\leqslant b_i\leqslant a_i).$$

Now consider the product $$(1+p_1+p_1^2+\dotsb+p_1^{a_1})(1+p_2+p_2^2+\dotsb+p_2^{a_2}) \dotsm(1+p_m+p_m^2+\dotsb+p_m^{a_m})$$ which when expanded will contain all the divisors of $n$ in a sum of the divisors of $n$. As each bracketed term has $(a_i+1)$ terms, these being the values $b_i$ can take, namely $0$ to $a_i$, this gives the number of divisors of $n$, $\sigma_0(n)$, alternatively written $s(n)$, by the product

$$\sigma_0(n)=\prod_{i=1}^m(a_i+1)=(a_1+1)(a_2+1)\dotsm(a_m+1).$$ The sum of the divisors of $n$, $\sigma(n)$ is then

$$ \sigma(n)=\prod_{i=1}^m(1+p_i+p_i^2+\dotsb+p_i^{a_i}) =\prod_{i=1}^m\frac{p_i^{a_i+1}-1}{p_i-1}=\prod_{i=1}^m\sigma(p_i^{a_i}). $$

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  • $\begingroup$ You could also think about how you can expand on the above to work out the sum of the $k$th powers of the divisors of $n$, $\sigma_k(n)$. $\endgroup$ – Daniel Buck Sep 30 '16 at 21:02
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$$\sigma(n) = \sum_{d \mid n} d$$

If $\gcd(n,m) = 1$ then there is bijection $(d,d') \to dd'$ between the couples of divisors $ d \mid n, d' \mid m$ and the divisors of $ nm$, and hence $$\sigma(n)\sigma(m) = (\sum_{d \mid n} d)(\sum_{d' \mid m} d') = \sum_{d \mid n, \ d' \mid m} dd' = \sum_{k \mid nm} k=\sigma(nm)$$

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  • $\begingroup$ Hi, i cant understand how the bijection between $(d,d')$ and $dd'$ is done. Can you please prove it in details ? Thank you $\endgroup$ – outlawoutlawz Oct 1 '16 at 5:56
  • $\begingroup$ @RamyTanios show that $(d,d') \mapsto dd'$ is bijective $\endgroup$ – reuns Oct 1 '16 at 15:08

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