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 $a,b,c,d$.

  • 1
    $\begingroup$ What do you mean by "evaluate"? $\endgroup$
    – Will Sawin
    Jun 7, 2018 at 22:37
  • $\begingroup$ Opening up the last factor $d(n-m)$ yields $$ \sum_{c\le X} \sum_{m\le X} d(m) \sum_{\substack{m<n\le X \\ n \equiv m \pmod c}} d(n), $$ which might be tractable.... $\endgroup$ Jun 8, 2018 at 0:48
  • $\begingroup$ @WillSawin: I mean an asymptotic formula, or lower and upper bounds of the right order of magnitude $\endgroup$ Jun 8, 2018 at 16:36

1 Answer 1


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.


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