How to prove $ \prod_{d|n} d= n^{\frac{\tau (n)}{2}}$

how to prove:

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

$\prod_{d|n} d$ is product of all of distinct positive divisor of $n$,

$\tau (n)$ is number (count)of all of positive divisor of $n$

Solution 1:

Hint: If $k$ divides $n$, then $\frac {n}{k}$ is an integer that divides $n$.

This allows you to pair up the divisors, to show that $$(\prod_{d \mid n} d )^2 = \prod_{d \mid n} n = n^{\tau(n)}.$$

Elaborated:

$$(\prod_{d \mid n} d )^2 = (\prod_{d \mid n} d ) ( \prod_{d \mid n} \frac {n}{d} ) = \prod_{d \mid n} d \times \frac {n}{d} = \prod_{d \mid n} n = n^{\tau(n)}.$$

Solution 2:

Note that positive divisors of $n$ go in pairs, as if $d$ is such a divisor, so is $n/d$.

However, some care must be taken when $n = k^2$ is a perfect square, because you have only one factor $k$ in the product. In this case $$ \prod_{d \mid n, d \ne k} = n^{\frac{\tau(n)-1}{2}}. $$ Now multiply both sides by $k = n^{1/2}$.

Solution 3:

suppose $\tau(n)=r (I)$,$d_1,...,d_r$ is all of distinc positive divisor $n$

$d_i|n$ thus $\frac n{d_i}|n$ so {$\frac n{d_1},...,\frac n{d_r}$}$\subset${$d_1,...,d_r$}

also we have: $d_i=\frac n{\frac n{d_i}}=\frac n{d_j}$ so

{$\frac n{d_1},...,\frac n{d_r}$}$=${$d_1,...,d_r$}

$$\prod_{d|n} d =d_1...d_r=\frac n{d_1}.....\frac n{d_r}$$ so $n^r=(d_1...d_r)^2$

$n^{\frac r2}=\prod_{d|n} d =d_1...d_r=n^{\frac {\tau(n)}2}$

Solution 4:

Note that for each divisor $d$ of $n$ there is divisor $d'=n/d$ such that $dd'=n$.

The result then follows at once if $n$ is not a square. I'll leave this case for the reader. :-)