Prove that $ a_{1}^{3}+a_{2}^{3}+\cdots+a_{l}^{3}=\left(a_{1}+a_{2}+\cdots+a_{l}\right)^{2} $

Question

Let $ d_{1}, d_{2}, \ldots, d_{l}$ be all positive divisors of a positive integer $n$. For each $i=1,2, \ldots, l$ denote by $a_{i}$ the number of positive divisors of $d_{i}$. Then $ a_{1}^{3}+a_{2}^{3}+\cdots+a_{l}^{3}=\left(a_{1}+a_{2}+\cdots+a_{l}\right)^{2} $

Solution. hint says, we have to use this but I am not getting why these both are true ???

$ \begin{aligned} a_{1}+a_{2}+\cdots+a_{l} &=\sum_{d \mid n} \tau(d)=\prod_{i=1}^{k}\left(1+\tau\left(p_{i}\right)+\cdots+\tau\left(p_{i}^{\alpha_{i}}\right)\right) \\ a_{1}^{3}+a_{2}^{3}+\cdots+a_{l}^{3} &=\sum_{d \mid n} \tau(d)^{3}=\prod_{i=1}^{k}\left(1+\tau\left(p_{i}\right)^{3}+\cdots+\tau\left(p_{i}^{\alpha_{i}}\right)^{3}\right) \end{aligned} $

And what are the $p_i$? Presumably having to do with prime power factorization of $n$ but this should be clarified in the problem. — halrankard, Jul 02 '20 at 14:46
Take an arbitrary element of the first product. What does it look like, and can you make that correspond to some divisor of $n$? Hopefully you can finish from there. — rogerl, Jul 02 '20 at 15:08
Does $d_1,...,d_l$ include the trivial divisors $1$ and $n$? — K.defaoite, Jul 02 '20 at 15:13
Related question: https://math.stackexchange.com/questions/1398613. — Batominovski, Jul 02 '20 at 16:57
@Ishan have you studied the method that I posted? It's quite straightforward as long as you can prove multiplicativity of tau and the multiplicativity of the sum of a multiplicative function over the positive divisors of the input. — Favst, Jul 03 '20 at 19:14

score 0 · Answer 1 · answered Jul 02 '20 at 15:50

See Proposition 1 in Sum of Cubes is Square of Sum by Barbeau and Seraj. Here's the quoted proof:

The proposition is clearly true for prime powers since for each prime $p$ and each natural $n$, the corresponding set of $p^{n-1}$ is $$\langle\tau(p^0), \tau(p^1), \ldots, \tau(p^{n-1})\rangle=\langle 1,2,\ldots,n\rangle.$$ $\tau$ is well-known to be a multiplicative function, meaning that if $a,b$ are relatively prime integers then $\tau(ab)=\tau(a)\tau(b).$ By elementary multiplicative number theory, the functions \begin{align*} f(n)&=\sum_{d\mid n}{\tau(d)},\\ F(n)&=\sum_{d\mid n}{[\tau(d)}]^3 \end{align*} must also be multiplicative. All we need now is that $F(n)=[f(n)]^2$. Since this is already true for prime powers, the rest follows from the prime factorization of $n$ and the multiplicative property of $f$ and $F$.

The result is attributed to Liouville. Disclosure: I'm one of the authors, and I have no issues with the proof in the paper being quoted here. See here for why the sum of a multiplicative function over divisors of the input is multiplicative. Of course, a multiplicative function to a positive integer exponent is also multiplicative.

Prove that $ a_{1}^{3}+a_{2}^{3}+\cdots+a_{l}^{3}=\left(a_{1}+a_{2}+\cdots+a_{l}\right)^{2} $

1 Answers1