Exercise 4.2.24* ($\sum_{d\mid n}d(d)^3 = (\sum_{d\mid n} d(d))^2$)

Question

Show that $\sum_{d\mid n}d(d)^3 = (\sum_{d\mid n} d(d))^2$ for all positive integers $n$.

Richard Ganaye · Accepted Answer

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\begin{proof} 
We want to show that
$$
\sum_{\delta \mid n}d(\delta)^3 = \left(\sum_{\delta \mid n} d(\delta)ight)^2.
$$
The product of multiplicative functions is a multiplicative function. Since $d$ is a multiplicative function, so is $d^3$. By Theorem 4.4, the function $F : \N^* 	o \Z$, defined by
$n \mapsto F(n) = \sum_{\delta \mid n}d(\delta)^3,$
is a multiplicative function.

By the same theorem, $ H : n \mapsto \sum_{\delta \mid n} d(\delta)$ is multiplicative, thus $G = H^2$ is multiplicative, i.e.
$$n \mapsto  G(n) = \left(\sum_{\delta \mid n} d(\delta)ight)^2$$
is multiplicative.

Now we compare $F(p^\alpha)$ and $G(p^\alpha)$, where $p$ is a prime number.
\begin{align*}
F(p^\alpha) &= \sum_{\delta \mid p^\alpha} d(\delta)^3\
&= \sum_{i=0}^\alpha d(p^i)^3\
&= \sum_{i=0}^\alpha (i+1)^3\
&= \sum_{j=1}^{\alpha +1} j^3\
&=\frac{(\alpha +1)^2(\alpha +2)^2}{4},
\end{align*}
and
\begin{align*}
G(p^\alpha) &= \left( \sum_{\delta \mid p^\alpha} d(\delta) ight)^2\
&=\left( \sum_{i=0}^\alpha (\alpha + 1) ight)^2\
&=\left( \sum_{j=1}^{\alpha +1}j ight)^2\
&=\left(\frac{(\alpha + 1)(\alpha+2)}{2} ight)^2.
\end{align*}
So $F(p^\alpha) = G(p^\alpha)$. Since $F,G$ are multiplicative functions, $F = G$, so
$$
\sum_{\delta \mid n}d(\delta)^3 = \left(\sum_{\delta \mid n} d(\delta)ight)^2.
$$
This is a consequence of the well known identity 
$$\sum_{j=1}^n j^3 = \left( \sum_{j=1}^n j ight)^2.$$
\end{proof}

Exercise 4.2.24* ($\sum_{d\mid n}d(d)^3 = (\sum_{d\mid n} d(d))^2$)

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Exercise 4.2.24* ($\sum_{d\mid n}d(d)^3 = (\sum_{d\mid n} d(d))^2$)

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