Let $d(n)$ denote the number of positive divisors of $n$. Is it known how to evaluate the sum


$$\displaystyle \sum\_{1 \leq m < n \leq X} d(m) d(n) d(n-m)?$$


A slightly more difficult question is if we change the height condition in the summation, to obtain the sum


$$\displaystyle \sum\_{\substack{1 \leq mn(n-m) \leq X \\ 1 \leq m < n}} d(m) d(n) d(n-m).$$


This is a generalization of the single variable case, where it is known how to evaluate sums of the form


$$\displaystyle \sum\_{1 \leq n \leq X} d(an + b) d(cn+d)$$


for fixed positive integers $a,b,c,d$. 



A sum of divisor functions
The first problem is completely solved in the paper:


Tim Browning - The divisor problem for binary cubic forms.
J. Théorie Nombres Bordeaux 23 (2011), 579-602.


The method is to change the order of summation to reduce to a lattice point counting problem. One controls the error term using a variant of Dirichlet's hyperbola method.


Similar methods should work for the second problem, as your region is homogeneous hence should be amenable to lattice point counting techniques.


