What does the product of all proper divisors equal to?

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

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Like Sivaram points out. If $n$ is not a perfect square, then every divisor $d_1$ can be paired with the divisor $\frac{n}{d_1}$, which is distinct from $d_1$; the product of these two is $n$. So the product of all divisors is equal to $n^k$, where $2k$ is the number of divisors. (Note that a positive integer has an odd number of distinct divisors if and only if it is a square).

If $n$ is a perfect square, then it will have $2k+1$ divisors for some $k$; the product will be $n^k\times\sqrt{n} = n^{k+\frac{1}{2}}$.

Either way, the answer is $n^{d(n)/2}$, where $d(n)$ is the number of divisors of $n$ (sometimes written $\sigma_0(n)$ or $\tau(n)$).

If you want the proper divisors, then you are excluding $n$ from the product, so the answer becomes $n^{(d(n)/2) – 1}$.

We are meant to calculate,

$$
\prod_{t|n}t, t\neq n
$$

We will calculate including $t=n$, and will divide the result finally by $n$.

Using fundamental theorem of arithmetic,

$$
n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}
$$

where $p_j$ is a prime number. Now if $p^a$ is an entry in the above, number of terms in above product containing $p^{a’}$ is $d(n)/(a+1)$, where $d(n)$ denotes the number of positive divisors of number $n$. Thus in above product, contribution out of prime $p$ is

$$
p^{(1+2+\cdots +a)*d(n)/(a+1)}=(p^a)^{d(n)/2}
$$

So,

$$
\prod_{t|n}t=(p_1^{a_1})^{d(n)/2}(p_2^{a_2})^{d(n)/2}\cdots (p_k^{a_k})^{d(n)/2}=n^{d(n)/2}
$$

Division by $n$ gives,

$$
n^{\frac{d(n)}{2}-1}
$$