2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ How do u define it if m/n is not an integer ? $\endgroup$ – user379195 May 9 '17 at 21:22
  • 5
    $\begingroup$ true when $m/n$ is an integer AND $\gcd(n, m/n) = 1$ $\endgroup$ – Will Jagy May 9 '17 at 21:25
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
4
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.