I am trying to prove or disprove the following statement:

$$\sum_{r|n} d(r^2) = d^2(n),$$

where $d$ is the number of divisors function.

Computing it for small numbers yields equality, so I at least think it's true. But I can't see how to prove it formally. I have tried working with prime factorisations, but finding equality between terms seems near-impossible at best. My best guess is there is a way to manipulate

$$\sum_{r\mid n} \sum_{k\mid r^2} 1$$

into a form which becomes the RHS, although my attempts at writing $n=ar$ and $r^2=bk$ for $a,b\in\mathbb{Z}$ become very messy when I attempt substitution.

Any assistance or hints would be greatly appreciated! Thank you!

  • $\begingroup$ What divisor function do you mean? just divisors? unique divisors? prime divisors? unique prime divisors? $\endgroup$ – Mastrem Oct 29 '16 at 10:31
  • $\begingroup$ Just divisors - for example $d(6) = 4$ since it has four distinct divisors. $\endgroup$ – Camille Oct 29 '16 at 10:35

Write $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$. Then \begin{align}\sum_{r|n} d(r^2)&=\sum_{0\leq i_1 \leq \alpha_1,\ldots,0\leq i_k \leq \alpha_k} d((p_1^{i_1}\cdots p_k^{i_k})^2)\\&=\sum_{0\leq i_1 \leq \alpha_1,\ldots,0\leq i_k \leq \alpha_k} d(p_1^{2i_1}\cdots p_k^{2i_k})\\&=\sum_{0\leq i_1 \leq \alpha_1,\ldots,0\leq i_k \leq \alpha_k}(2i_1+1)\cdots(2i_k+1)\\&=\left(\sum_{0\leq i_1 \leq \alpha_1}(2i_1+1)\right)\cdots \left( \sum_{0\leq i_k \leq \alpha_k}(2i_k+1)\right)\\&=\left(2\frac{\alpha_1(\alpha_1+1)}{2}+(\alpha_1+1)\right)\cdots \left( 2\frac{\alpha_k(\alpha_k+1)}{2}+(\alpha_k+1)\right)\\&=(\alpha_1+1)^2\cdots(\alpha_k+1)^2\\&=d(n)^2.\end{align}

  • $\begingroup$ Thank you! Just to clarify, the limits of summation in the fifth line should read $0\leq i_k \leq a_k$, yes? $\endgroup$ – Camille Oct 29 '16 at 11:39
  • $\begingroup$ Yes, that's right. $\endgroup$ – SMJK Oct 29 '16 at 11:42

The divisor counting function $d$ is multiplicative, so it is straightforward to verify that both $d(n^2)$ and $d^2(n)$ are multiplicative. Furthermore, by a fundamental theorem of multiplicative functions, $\sum_{r|n} d(r^2)$ is also multiplicative. Hence, because both sides of the statement are multiplicative, to establish that the statement is true for all positive integers $n$, it suffices to prove that it is true for all prime powers $p^\alpha$:

\begin{align} \sum_{r|p^\alpha} d(r^2) & = d(1^2) + d(p^2) + d(p^4) + \cdots + d(p^{2\alpha}) &&\text{positive divisors of } p^\alpha \text{are } 1, p, p^2, \dots, p^\alpha\\ & = 1 + 3 + 5 + \cdots + 2\alpha + 1 && d(p^i) = i + 1\\ & = (\alpha + 1)^2 &&\text{sum of the arithmetic series}\\ & = d^2(p^\alpha) && i + 1 = d(p^i) \end{align}


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