Prove that if $n \in \mathbb{N}$, then $\sum_{d|n}{(d(n))^3}=(\sum_{d|n}{d(n)})^2$ where $d(n)$ is the divisor function.

Prove that if $n \in \mathbb{N}$, then $$\sum_{d|n}{(d(n))^3}=(\sum_{d|n}{d(n)})^2$$ where $d(n)$ is the divisor function. I have know only information is $d(n)=\sum_{d|n}{1}$.

Solutions Collecting From Web of "Prove that if $n \in \mathbb{N}$, then $\sum_{d|n}{(d(n))^3}=(\sum_{d|n}{d(n)})^2$ where $d(n)$ is the divisor function."

Outline: Let $f(n)$ be the function on the left, and $g(n)$ the function on the right.

Both $f$ and $g$ are multiplicative.

So we only need to verify that the equality holds when $n$ is a power of a prime.
Multiplicativity then atutomatically gives us the rest.

The number $p^k$ has $k+1$ divisors: in symbols, $d(p^k)=k+1$.

So we want to prove that for any $m$,
$$\sum_0^m (k+1)^3 =\left(\sum_0^m (k+1)\right)^2.$$
Look up the formula for the sum of the first $q$ cubes.