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?
 A: If $a$ and $b$ are coprime, every odd divisor of $ab$ can be written in exactly one way as the product of an odd divisor of $a$ and and odd divisor of $b$, and every such product is an odd divisor of $ab$.  So $f(ab) = f(a) f(b)$.
A: 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.
A: 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, $f(mn) = f(m)f(n)$.
