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 $d(n)=\sum_{d|n}{1}$.
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See also http://math.stackexchange.com/a/216698/589. – lhf Aug 30 '13 at 14:23
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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.
André Nicolas
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Yes, any integer $\ge 0$. You may be more familiar with $1^3+2^3+3^3+\cdots +n^3=\frac{n^2(n+1)^2}{4}$, which says the same thing. If you don't know the result, it can be proved by induction. – André Nicolas Aug 30 '13 at 15:25