Exercise 4.2.5 ($\prod_{d \mid n} d= n ^{d(n)/2}$)

Question

Prove that $\prod_{d \mid n} d= n ^{d(n)/2}.$

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $n = p_1^{a_1} \cdots p_l^{a_l}$ the canonical decomposition of $n$. Then any divisor $d$ of $n$ is of the form
$$d = p_1^{i_1}\cdots p_l^{i_l},\qquad 	ext{where } (i_1,\ldots,i_l) \in [\![0,a_1]\!] 	imes [\![0,a_2]\!] 	imes \cdots  	imes[\![0,a_l]\!] .$$

We use the rule 
\begin{align}
\prod_{i=0}^n ka_i = k^{n+1}\prod_{i=0}^n a_i,
\end{align}
where $k$ is a constant independent of $i$.

Using several times the rule (1), we obtain
\begin{align*}
\prod_{d \mid n} d &= \prod_{(i_1,\ldots,i_l) \in [\![0,a_1]\!] 	imes [\![0,a_2]\!] 	imes \cdots  	imes[\![0,a_l]\!]} p_1^{i_1} \cdots p_l^{i_l}\
&=\prod_{i_1 = 0}^{a_1} \prod_{i_2 = 0}^{a_2}  \cdots \prod_{i_l = 0}^{a_l} p_1^{i_1} p_2^{i_2}\cdots p_l^{i_l}\
&= \prod_{i_1 = 0}^{a_1}  (p_1^{i_1})^{(a_2+1)\cdots (a_l +1)} \prod_{i_2 = 0}^{a_2}  \cdots \prod_{i_l = 0}^{a_l} p_2^{i_2} \cdots p_l^{i_l}\
&= \prod_{i_1 = 0}^{a_1}  p_1^{i_1 (a_2+1)\cdots (a_l +1)} m. &(	ext{where } m =  \prod_{i_2 = 0}^{a_2}  \cdots \prod_{i_l = 0}^{a_l} p_2^{i_2} \cdots p_l^{i_l})\
&= m^{a_1+1} \prod_{i_1 = 0}^{a_1}  p_1^{i_1 (a_2+1)\cdots (a_l +1)}\
&=m^{a_1+1} p_1^{(a_2+1)\cdots (a_l +1) \sum_{i_1= 0}^{a_1} i_1}\
&=m^{a_1+1}  p_1^{\frac{a_1(a_1+1)(a_2+1) \cdots (a_l + 1)}{2}}\
&=m^{a_1+1} p_1^{a_1 d(n)/2}.
\end{align*}
Here $m = \prod_{i_2 = 0}^{a_2}  \cdots \prod_{i_l = 0}^{a_l} p_2^{i_2} \cdots p_l^{i_l}$ is prime to $p_1$, so $m^{a_1+1} \wedge p_1 = 1$. Therefore
$$
u_{p_1}\left( \prod_{d \mid n} d ight) =\frac{a_1 d(n)}{2}.$$
Since all the $p_i$ play a symmetric role, we obtain similarly
$$
u_{p_i}\left( \prod_{d \mid n} d ight) =\frac{a_i d(n)}{2}\qquad (1 \leq i \leq l).$$
Since $p_1,\ldots,p_n$ are the only divisors of $\prod_{d \mid n} d$,
\begin{align*}
\prod_{d \mid n} d &= p_1^{a_1 d(n)/2} p_2^{a_2 d(n)/2}\cdots p_l^{a_l d(n)/2}\
&=\left(  p_1^{a_1} p_2^{a_2}\cdots p_l^{a_l}ight)^{d(n)/2} \
&=n^{d(n)/2}.
\end{align*}
\end{proof}

Note: $d(n)$ is not always even, but $d$ is odd if and only if $n$ is a square (see Problem 12), so $n^{d(n)/2}$ is an integer.

Exercise 4.2.5 ($\prod_{d \mid n} d= n ^{d(n)/2}$)

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Exercise 4.2.5 ($\prod_{d \mid n} d= n ^{d(n)/2}$)

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