My question is the same as with the title. Kindly help me prove the statement below if it is true.

What I tried: For specific value of $m$ and $n$ the equality seems to hold.

Does $\sigma(\frac{m}{n})=\frac{\sigma(m)}{\sigma(n)}$ where $\sigma$ is the sum of divisor function?

If not what are the conditions for $m$ and $n$ such that the above condition is true.

Thanks a lot.

  • 1
    $\begingroup$ How do u define it if m/n is not an integer ? $\endgroup$ – user379195 May 9 '17 at 21:22
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    $\begingroup$ true when $m/n$ is an integer AND $\gcd(n, m/n) = 1$ $\endgroup$ – Will Jagy May 9 '17 at 21:25
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    $\begingroup$ In general, no. This is $\sigma(a)\sigma(b) = \sigma(ab)$. True if $a,b$ are relatively prime, but not true in general. $\endgroup$ – GEdgar May 9 '17 at 21:25
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    $\begingroup$ For a counterexample, take $m=4,n=2$. $\endgroup$ – GEdgar May 9 '17 at 21:26
  • $\begingroup$ Thanks for your comments. Specially to Sir Will Jagy. $\endgroup$ – Jr Antalan May 9 '17 at 21:46

let $N(k)=k$ for all $k$ so that $N$ is a completely multiplicative function. by a well-known result in the elementary theory of arithmetic functions, this implies that $$ \sigma(n)=\sum_{d|n}N(d) $$ is also a multiplicative function. so, as Will Jagy pointed out, if $n|m$ and $\gcd(n,\frac{m}{n}=1)$ we have $$ \sigma(m)=\sigma(n\frac{m}{n})=\sigma(n)\sigma(\frac{m}{n}) $$ from which the result stated follows

  • $\begingroup$ Nice answer. Now I know. Till next time. $\endgroup$ – Jr Antalan May 9 '17 at 21:50

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