Prove or disprove: $ \sum_{b \vee d = x} \tau(b) \tau(d) = \tau(x)^3$ Can somebody prove or disprove?  Let $\tau$ be the divisors function, so that $\tau(6) = \#\{ 1,2,3,6\} = 4$
$$ \sum_{b \vee d = x} \tau(b) \tau(d) = \tau(x)^3$$
Here I am using $b \vee d = \mathrm{lcm}(b,d)$ since it is the join of two numbers in the multiplicative lattice $(\mathbb{N}, \times)$.

This statement seems to be true.  Let's try $x = 6$.  The left hand side is:
$$ 2\times\tau(6)\big[\tau(1)+\tau(2)+\tau(3)\big]+\tau(6)^2+
   \tau(3)\big[\tau(2)+\tau(6)\big]+\tau(2)\big[\tau(3)+\tau(6)\big] = 64$$
and indeed $\tau(6)^3 = 64$.

One option is to try and take advantage of Dirichlet series identities:
\begin{eqnarray}
 \zeta(s)^2 &=& \sum \frac{\tau(n)}{n^s} \\
 \frac{\zeta(s)^4}{\zeta(2s)} &=& \sum \frac{\tau(n)^2}{n^s} \\
\end{eqnarray}
Wikipedia doesn't mention and identity for $\tau(n)^3$.  This is just one possible proof attack.
 A: If $x = p^n$, then we have $b\vee d = x$ if and only if $b = x$ or $d = x$, so
$$\sum_{b\vee d = p^n} \tau(b)\tau(d) = 2\tau(p^n)\sum_{k = 0}^n\tau(p^k)  - \tau(p^n)^2 = 2(n+1)\frac{(n+1)(n+2)}{2} - (n+1)^2 = (n+1)^3.$$
The right hand side, $\tau(x)^3$ is multiplicative, and the left hand side
$$\sum_{b\vee d = x} \tau(b)\tau(d)$$
is also multiplicative, since for $\gcd(x,y) = 1$ we have
$$(b_1b_2)\vee (d_1d_2) = xy \iff (b_1\vee d_1 = x) \land (b_2\vee d_2 = y)$$
for divisors $b_1,d_1$ of $x$ and $b_2,d_2$ of $y$.
Thus indeed
$$\sum_{b\vee d = x} \tau(b)\tau(d) = \tau(x)^3.$$
A: This follows from the identity
$$\left( \sum_{d \mid x} \tau(d) \right)^2 = \sum_{d \mid x} \tau(d)^3$$
as follows. We have
$$\begin{align*}
\sum_{n \mid x} \tau(n)^3
&= \left( \sum_{d \mid x} \tau(d) \right)^2 \\
&= \sum_{b, d \mid x} \tau(b) \tau(d) \\
&= \sum_{n \mid x} \sum_{lcm(b, d) = n} \tau(b) \tau(d)
\end{align*}$$
By Möbius inversion, we may cancel the sum $\sum_{n \mid x}$ and keep only the term for $n = x$.

Similarly (*), the identity
$$\sum_{d \leq x} d^3 = \left( \sum_{d \leq x} d \right)^2$$
implies
$$x^3 = \sum_{\max(b, d) = x} bd$$
(*) To make the analogy complete, you can view $\mathbb N$ as a monoid of Beurling integers for the operation of addition, with a single Beurling prime, $1$. With respect to this prime, we  have $\tau(n) = n+1$ and the Möbius function satisfies $\mu(0) = 1$, $\mu(1) = -1$ and $\mu(n) = 0$, $n > 1$. Now clearly
$$\sum_{d \mid x} \tau(d)^3 = \sum_{d \leq x+1} d^3$$
and so on.
