Proof of a sum of positive divisors Let $n$ be an integer greater than zero. Prove
$$(\sum_{d|n}v(d)){}^{2}=\sum_{d|n}(v(d))^{3}$$
where $v(d)$ is the number of positive divisors of $n$.
I'll outline what my problem is. I write $n= p_{1}^{a_{1}}, p_{2}^{a_{2}}, \cdots , p_{m}^{a_{m}}$. Then I write out each divisor $d_{1}, d_{2}, ..., d_{(a_{1} + 1)(a_{2}+1)...(a_{m} + 1)}$ for $n$. Then
$$(\sum_{d|n}v(d)){}^{2}=(v(d_{1})+v(d_{2})+\ldots+v(d_{(a_{1}+1)(a_{2}+1)\cdots(a_{m}+1)}))^{2},
 $$
but now how do I simplify anything? I don't know how many divisors of EACH $d_{i}$ there are. $d_{i}$ might be prime, it might have 3 divisors--I don't know. That depends on the prime factorization of $d_{i}$. So now I have to write each $d_{i}$ in its very own prime factorization, so I'll have $(a_{1}+1)(a_{2}+1)\cdots(a_{m}+1)$ prime factorizations written out. That's just silly. I realize $v(d)$ is multiplicative but so what? I still have to actually know how many divisors each divisor will have, which means I have to write out each divisor in its own prime factorization.
 A: Quickly show $v$ is multiplicative:
Let $a, b$ be relatively prime integers, so $a=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{m}^{a_{m}}$ and $b=p_{1'}^{b_{1}}p_{2'}^{a_{2}}\cdots p_{n}^{a_{n}}
 $. They have no primes in common as we assume they are relatively prime.
Then $$v(ab) = v((p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{m}^{a_{m}})(p_{1'}^{b_{1}}p_{2'}^{a_{2}}\cdots p_{n}^{a_{n}}))$$
$$=  v(p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{m}^{a_{m}}p_{1'}^{b_{1}}p_{2'}^{a_{2}}\cdots p_{n}^{a_{n}})$$
and by a properties previously established in my textbook
$$=(a_{1}+1)(a_{2}+1)\cdots(a_{m}+1)(b_{1}+1)(b_{2}+2)\cdots(b_{n}+1)$$
$$=\left((a_{1}+1)(a_{2}+1)\cdots(a_{m}+1)\right)\left((b_{1}+1)(b_{2}+2)\cdots(b_{n}+1)\right)$$
$$v(a)v(b).$$
I realize this leaves out a lot but for purposes of my question and given the information in the book that has already been established, it would suffice for my assignment (I believe).
So now this means $v(d)$ is completely determined by its values on prime powers, and so we can assume $n$ is a prime power. Let $n = p^{a}$ for some positive integer $a$. Then all the divisors of $n$ are $\left\{ 1,p,p^{2},p^{3},\ldots,p^{a}\right\}.$
$$(\sum_{d|n}v(d)){}^{2}=(\sum_{d|p^{a}}v(d)){}^{2}$$
$$= v(1) + v(p) + v(p^{2}) + ... + v(p^{a})$$
$$=(1+(1+1)+(2+1)+...+(a+1))^{2}$$
$$=(1+2+3+...+a+1)^{2}$$
$$=\left(\sum_{k=0}^{a}k+1\right)^{2}$$
and by Faulhaber's formula, we have
$$=\sum_{k=0}^{a}\left(k+1\right)^{3}$$
$$=1^{3}+ 2^{3} + 3^{3} + ... + (a+1)^{3}$$
$$=1 + (1+1)^{3} + (2+1)^{3} + ... (a+1)^{3}$$
So by definition of $v$,
$$=v(1) + v(p^{1})^{3} + v(p^{2})^{3} + ... + v(p^{a})^{3}$$
$$=\sum_{d|p^{a}}\left(v(d)\right)^{3}$$
and since $n=p^{a}$,
$$\sum_{d|n}\left(v(d)\right)^{3}\blacksquare
 $$
