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?

Firstly, multiplicative only means for relatively prime integers $m$ and $n$. The integers $m$ and $n$ can be expressed as a product of primes: $m = 2^{a_1} {p_2}^{a_2} {p_3}^{a_3} ... {p_n}^{a_n}$. $n = 2^{b_1}{q_2}^{b_2} {q_3}^{b_3} ... {q_n}^{b_n}$.Note that $p_i \not= q_i$ for any $i$ because m and n are relatively prime [$a_1$ or $b_1$ is $0$, and all other powers are nonzero]. 
We thus can consider the set $p_2, p_3 .... p_n, q_2, .... q_n$ to obtain all odd factors. (We removed factors that are a multiple of 2). It will be easier if we rename the elements of the set, calling the $i$th element as $s_i$. So we must find the power set of the set $\{ s_1, s_2, ......... s_{2n-2}\}$. [the null set takes the role of $1$ which is not a product of primes]. So $f(mn) = (2n-1)C0 + (2n-2)C 1 + (2n-2)C2 + ... (2n-2) C (2n-2) = (1+1)^{2n-2} = 2^{2n-2}$.
Okay. Now, $f(m) = 2^{(n-1)}$ and $f(n) = 2^{(n-1)}$ [I hope you can see why, by applying the same technique], and we have shown that $f(mn) = 2^{2n-2}$. Thus, $f(mn) = f(m)f(n)$.
