arithmetic relations between divisors of a natural number

Question

For n,d $\epsilon$ $\mathbb N$ prove that

$\prod_{d|n}$ $d^2$ = $n^{d(n)}$

where d(.):$\mathbb N$ $\to $ $\mathbb N$ is defined by d(n):= $\sum_{d|n} 1$

So d(n) is the number of positive divisors of n and the left hand side is product of squares of the positive divisors of n.

One hint I was given was that if d is a positive divisor of n, then so is $\frac{n}{d}$

Answer 1

Let $n\in\mathbb{N}$, $D=\lbrace d_0=1,d_1, d_2,\ldots,d_l=n\rbrace$ be the set of divisors of $n$, $d(n)=l+1$.

Note that: $d$ is divisor of $n$ $\Leftrightarrow$ $\dfrac{n}{d}$ is divisor of $n$.
It means that $d_i\cdot d_{l-i}=n, \forall i=0, 1,\ldots,l$.

So $\prod_{d|n}d^2=d_0\cdot d_0\cdot d_1\cdot d_1\cdot\ldots\cdot d_l\cdot d_l= (d_0\cdot d_l)\cdot(d_1\cdot d_{l-1})\cdot\ldots(d_l\cdot d_0)= \underbrace {n\cdot\ldots \cdot n}_{l+1} = n^{l+1} = n^{d(n)}$

Answer 2

Let the divisors of $n$ be $d_1,d_2,\cdots, d_k$ where $k=d(n)$. You realise right, that $d_1$ and $\frac{n}{d_1}$, both being divisors of $n$ occur in this list?

There can be two cases, one, when $d_i=\frac{n}{d_i}$ for some $i\in \big\{1,2,\cdots,k\big\}$, so that $d_i=\sqrt{n}=n^{\frac{1}{2}}$.

So the product $\Pi_{d\mid n} d = d_i \times (d_1 .\frac{n}{d_1})\times \cdots \times (d_{i-1}.\frac{n}{d_{i-1}}) \times (d_{i+1}.\frac{n}{d_{i+1}}) \cdots \times (d_k. \frac{n}{d_k})= n^{\frac12}\times n \times \cdots \times n$ It is easy to see that there are $k-1$ factors except $d_i$ in the above product, and hence $n$ appears $\frac{k-1}{2}$ times. So we have $$ \Pi_{d\mid n} d = n^{\frac12+\frac{k-1}{2}}=n^{\frac{k}2}$$, so the required product asking for the square of the square of the above product is simply $n^k=n^{d(n)}$ (since $k=d(n)$).

The case where there is no $i\in \big\{1,2,\cdots k \big\}$ such that $d_i=\frac{n}{d_i}$ is even easier. In that case there will just be $\frac{k}{2}$ pairs of factors each of whose product is $n$, giving the same result..