0

I am trying to solve the following problem.

Prove that: $$\sum_{t\mid n} d(t)^3= \left(\sum_{t\mid n}d(t)\right)^2$$

I know that there exists the following identity: $$\sum i^3=\left (\sum i\right)^2$$

What is this identity called? Does anyone have an approach to this problem?

Why can't I just substitute $d(t)$ for $i$?

I have thought about using $\sigma_\alpha(p^\alpha)$ but I am unsure of the correct way to proceed. This is problem $2.12$ from Apostol's Analytic Number Theory.

Snaw
  • 4,074
Prime
  • 869
  • https://en.m.wikipedia.org/wiki/Squared_triangular_number – Bumblebee Feb 15 '22 at 23:49
  • 5
    What is $d(t)?$ – Thomas Andrews Feb 15 '22 at 23:53
  • Use the weak multiplicativity of the expressions on both sides of your desired equality to reduce to the prime-power case, where the other identity gives the result... :) – paul garrett Feb 16 '22 at 00:08
  • See also https://math.stackexchange.com/questions/216673/sum-k-1n-a-k3-left-sum-k-1n-a-k-right2 – lhf Feb 16 '22 at 00:09
  • For your third question, try $n=2^N$. – lhf Feb 16 '22 at 00:14
  • 1
    Dirichlet convolution is defined on page 29. Theorem 2.6 on page 29 says that the product is commutative and associative. On page 35, Thm 2.14 says the Dirichlet product of multiplicative funtions is also multiplicative. What I am not seeing spelled out is the simplest case: if $f(n)$ is (number-theoretic) multiplicative, then $g(n) = \sum_t|n f(t) $ is also multiplicative. From what I can see, this is the same as $f \ast 1,$ where the constant function $1$ is multiplicative. On page 38 they say your divisor sum functions $\sigma_\alpha$ are multiplicative. They really do. – Will Jagy Feb 16 '22 at 01:06
  • 1
    damn it; $g(n) = \sum_{t|n} ; f(t) $ – Will Jagy Feb 16 '22 at 01:16
  • 1
    In the "identity" you mention, the summation is going from $i=1$ to $n$. In your problem, the summation is taking place over the divisors of $n$. So it's not the same. – Ted Feb 16 '22 at 02:49
  • 1
    @Ted, sure, they're not the same, but the first identity implies the second, with $n=2^N$ – lhf Feb 17 '22 at 00:02

0 Answers0