TITLE: Proof of $\sum_{d|n} {\tau}^3(d)=\left(\sum_{d|n}{\tau}(d)\right)^2$ (not standard proof)
QUESTION [9 upvotes]: I am trying to prove that

$\sum_{d|n} {\tau}^3(d)=\left(\sum_{d|n}{\tau}(d)\right)^2$

without using the fact that $\tau$ is multiplicative and products/sums of multiplicative funcions are also multiplicative (here you can find multiple arguments that show this result using such properties).
My approach was the following:

LHS counts the cardinality of the following set:
$$\bigcup_{d| n}\big\{(x,y,z,d)\ /\ x,y,z|d \big\}  $$
and RHS counts this one
$$\big\{(x,d,y,e)\ / \ x|d; \ y|e; \ d,e|n\big\}$$

I tried to establish an explicit bijection between those sets, but I already spent enough time and my biggest efforts on it, unsuccesful. Any idea or hint you can give me?
Thanks in advance!

REPLY [4 votes]: Edit: the following is wrong, because Yoneda's lemma can't be applied to hom functors restricted to subcategories without some conditions (which aren't attained here). But in the spirit of combinatorial species, I do think category theory is the correct setting in which to explore the question of natural bijections. I have some new back-of-the-envelope ideas I'll look at later.

I submit that there is no natural bijection between the two.
Consider the finite posets $\Lambda=\{x,y,z,d\}$ with $x,y,z\le d$ and $\Gamma=\{x,d\}\cup \{y,e\}$ with $x\le d$ and $y\le e$ and no other relations. These are objects in the category $\sf Pos$ of posets. Our equation states that $\hom_{\sf Pos}(\Lambda,I)$ and $\hom_{\sf Pos}(\Gamma,I)$ are isomorphic in the category $\sf Set$ of sets for any finite interval $I$ of $(\Bbb N,|)$ - meaning the positive naturals ordered by divisibility. Note that every finite interval $I$ of $(\Bbb N,\mid )$ looks like $[a,b]\cong[1,b/a]$.
Now, these two homsets are only equal when we're considering these intervals $I$, so let $\cal L$ be the category of posets isomorphic to an interval $[1,n]$ of $(\Bbb N,\mid)$. Interpret $A=\hom_{\sf Pos}(\Lambda,-)$ and $B=\hom_{\sf Pos}(\Gamma,-)$ as functors $A,B:{\cal L}\to\sf Set$. Is there a natural isomorphism $A\cong B$?
If there were, then $A$ and $B$ would have the same symmetry, i.e. ${\rm Aut}(A)\cong{\rm Aut}(B)$. However according to Yoneda's lemma ${\rm Aut}(A)={\rm Aut}(\Lambda)=S_3$ and ${\rm Aut}(B)={\rm Aut}(\Gamma)=C_2$.