Prove that a function $f(n)$ counting the number of odd divisors multiplicative How can I show that $f(n)$ is multiplicative, where $f(n)$ represents the number of the divisors of n in the form $2k + 1$? 
I'm studying algebra and I came across some questions on multiplicative functions (which should be number theory though), but there is any worked example of such proof. Can you help?

First note $f(1)=1$, then let compute $f(2^k)= 1$.
Note that the total number of divisors function is $\tau(n)$ which is multiplicative, so considering if $n=2^km$ with $m$ odd, we can see that $f(n)=f(m)$, so
$$f(2^km)={\tau(n)\over \tau(2^k)}=\tau(m)$$
and we know $\tau$ is multiplicative.
