ELI5: Explanation of why the product of divisors of a natural $n$ is always $n^{\frac{d(n)}{2}}$ where $d(n)$ gives the number of divisors

Question

I tried reading the answers on Product of Divisors of some $n$ proof and unfortunately couldn't understand them. The accepted answer mentions that

${\displaystyle\prod_\limits{d|n}{d^2}=\prod_\limits{d|n}{d\frac{n}{d}} = \prod_\limits{d|n}{n}=n^{d(n)} }$ where $d(n)$ gives the number of divisors of a number $n$ over $\mathbb N$,

which I assume proves something about the product of (every divisor of $n$ squared) being equal to $n^{d(n)}$. But I still don't get how this is related to the conclusion.

Which part do you find confusing? One of the equalities or just the conclusion? — cansomeonehelpmeout, Dec 13 '18 at 14:30
The equality between the first two $\prod$s as well as how it connects to the conclusion. — Nameless King, Dec 13 '18 at 14:33
The main idea is to pair up the divisors. For every divisor $d$ of $n$, the number $n/d$ is also a divisor of $n$. — SmileyCraft, Dec 13 '18 at 14:34
The part you seem to be missing is that the product of (every divisor of n, squared) is equal to (the product of every divisor of n), squared. Taking square roots of both sides gives the result. — Andrew Dudzik, Dec 13 '18 at 15:10

score 3 · Answer 1 · answered Dec 13 '18 at 14:31

List the factors in increasing order, then again in decreasing order, then multiply the first number in each list, the second etc. There are $d(n)$ calculations that each return $n$, so the overall product is $n^{d(n)}$. The product of $n$'s factors is just the square root of this, since we listed the factors twice.

cansomeonehelpmeout · Accepted Answer · 2018-12-13T14:59:17.890

2

Bearer of the curse, first $d$ is a divisor of $n\iff$ $\frac{n}{d}$ is a divisor of $n$. This is because $d\cdot\frac{n}{d}=n$, and $\frac{n}{d}$ is an integer. This would mean that $$\prod_{d\mid n} d=\prod_{d\mid n} \frac{n}{d}\tag{1}$$ because both products run over the same divisors just in opposite order. Multiplying both sides of $(1)$ by $\prod_{d\mid n} d$ gives $$\prod_{d\mid n} d^2=\prod_{d\mid n} d\prod_{d\mid n} \frac{n}{d}=\prod_{d\mid n} n\tag{2}$$

For the last equality you get an $n$ for each divisor $d$. There are $d(n)$ divisors, we get

$$\prod_{d\mid n} d^2=n^{d(n)}\tag{3}$$

Now, $\prod_{d\mid n} d^2=\left(\prod_{d\mid n} d\right)^2$ so taking the square-root of both sides of $(3)$ yields $$\prod_{d\mid n} d=n^{\frac{d(n)}{2}}$$

edited Dec 13 '18 at 14:59

answered Dec 13 '18 at 14:51

cansomeonehelpmeout

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Haha, did you start your answer with a Dark Souls II reference? ^^ – Jun 23 '21 at 23:54
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@DullandWitlessBoi Yes, indeed! – cansomeonehelpmeout Jun 24 '21 at 07:48
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Absolutely love the reference. I'm about to clear the Dragon Shrine for the first time. – Jun 24 '21 at 22:35
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@DullandWitlessBoi Good luck! Don’t you dare go hollow! – cansomeonehelpmeout Jun 25 '21 at 14:57
Lol, why thank you – Jun 26 '21 at 01:18

ELI5: Explanation of why the product of divisors of a natural $n$ is always $n^{\frac{d(n)}{2}}$ where $d(n)$ gives the number of divisors

2 Answers2