## Product of divisors

Given the number of all positive divisors of an unknown positive integer and the product of those divisors, how can one find the unknown integer?

[SOLVED]

#### quasi solved this puzzle:

Theorem. Let $$n$$ be a positive integer. Let $$k$$ be the number of all positive divisors of $$n$$. Let $$m$$ be the product of all positive divisors of $$n$$. Then $$n = m^{\frac{2}{k}}$$.

Proof. If $$n$$ is not a perfect square, the $$k$$ positive divisors of $$n$$ can be grouped in $$\frac{k}{2}$$ pairs $$(a, b)$$ with $$a \lt b$$ and $$ab = n$$. It follows that the product of the factors of n is $$n^{\frac{k}{2}}$$.

If $$n$$ is a perfect square, the $$k - 1$$ positive integer factors of $$n$$ other than $$\sqrt{n}$$ can be grouped in $$\frac{k - 1}{2}$$ pairs $$(a, b)$$ with $$a \lt b$$ and $$ab = n$$. The product of those factors is $$n^{\frac{k - 1}{2}}$$, hence the product of all the factors is $$n^{\frac{k - 1}{2}} \sqrt{n}$$, which simplifies to $$n^{\frac{k}{2}}$$.

In both cases, $$m = n^\frac{k}{2}$$, hence $$n = m^\frac{2}{k}$$.

#### Gabriel Elpers solved this puzzle:

Suppose there are $$D$$ divisors of the natural number $$N$$ and that $$P$$ is the product of all the divisors. For every divisor $$d$$ there is a unique quotient $$q$$ such that $$dq = N$$. The product of all divisors and their quotients is $$P^2$$ since each divisor appears in this expression exactly twice. Rearranging the factors so that each divisor is paired with its quotient, we see that $$P^2 = N^D$$. Therefore, $$N = P^{\frac{2}{D}}$$.

#### Siddhesh Chaubal solved this puzzle:

Consider the case in which the number of divisors of the positive integer $$n$$ is even. Let it be $$2k$$ where $$k$$ is a positive integer. Then the product of all divisors is $$n^k$$, since each divisor of $$n$$ has corresponding to it another divisor of $$n$$ such that their product is $$n$$. Since there are $$k$$ such pairs, their product is $$n^k$$. Thus the $$k$$th root of the product is $$n$$.

Now consider the case in which the number of divisors is odd. Let it be $$2k + 1$$ where $$k$$ is a non-negative integer. Then, the number is a perfect square. The product of all divisors in this case is $$n^k \sqrt{n}$$, i.e. $$\sqrt{n}^{2k + 1}$$. To find $$n$$, take the $$(2k + 1)$$th root of the product and square it.

#### Gnanasenthil G solved this puzzle:

Theorem. Let $$n$$ be a positive integer, let the number of all positive divisors of $$n$$ be $$k$$, and let the product of all positive divisors of $$n$$ be $$m$$. Then $$m = n^{\frac{k}{2}}$$.

Proof. Let $$A$$ be the set of all positive divisors of $$n$$. When a positive integer is divided by its divisor it gives a quotient which is another divisor of the positive integer.

If $$n$$ is a perfect square, there is an element $$x \in A$$ such that $$x^2 = n$$, so this element doesn't pair with a different element in $$A$$ such that their product is $$n$$. In this case, $$|A|$$ is an odd number since for every other element in $$A$$, there exists another element in $$A$$ such that their product is $$n$$. We have $$\sqrt{n}$$ and $$\frac{k - 1}{2}$$ pairs of numbers in $$A$$ such that the product of the numbers in each pair is $$n$$. Thus, the product of all elements in $$A$$ is

\begin{equation*} m = \sqrt{n} \cdot n^{\frac{k - 1}{2}} = n^{\frac{k}{2}}. \end{equation*}

If $$n$$ is not a perfect square, $$|A|$$ is even since each element in $$|A|$$ has a pair in $$A$$ such that their product is $$n$$. In this case, we have $$\frac{k}{2}$$ pairs of numbers in $$A$$ such that the product of the numbers in each pair is $$n$$. Thus, the product of all elements in $$A$$ is

\begin{equation*} m = n^{\frac{k}{2}}. \end{equation*}

#### Nilabja Ray solved this puzzle:

Let $$n$$ be the positive integer, $$k$$ be the number of all positive divisors of $$n$$ and $$m$$ be the product of those divisors.

Unless $$n$$ is a perfect square, $$k$$ is even, i.e. $$k = 2l$$ for a positive integer $$l$$. There are $$l$$ pairs of divisors and the product of the two numbers in each pair is $$n$$. Therefore, the product of all $$l$$ pairs is

\begin{equation*} m = n^l = n^{\frac{k}{2}}. \end{equation*}

If $$n$$ is a perfect square, $$k$$ is odd, i.e. $$k = 2l + 1$$ for a non-negative integer $$l$$. One of the divisors is $$\sqrt{n}$$ and there are $$l$$ pairs of divisors and the product of the two numbers in each pair is $$n$$. Therefore

\begin{equation*} m = n^{l + \frac{1}{2}} = n^{\frac{k}{2}}. \end{equation*}

#### Tim Ophelders solved this puzzle:

Let $$D$$ be the unknown set of divisors of an unknown positive integer $$A$$. The goal is to find $$A$$, given $$N = |D|$$ and $$P = \prod D$$.

Let $$D_\lt$$, $$D_\gt$$ and $$D_=$$ be the sets of numbers in $$D$$ smaller, greater and equal to $$\sqrt A$$, respectively. Their union is again $$D$$.

$D_\lt \cup D_\gt \cup D_= = D.$

Note that there is a bijection between $$D_\lt$$ and $$D_\gt$$.

$d \in D_\lt \Longleftrightarrow \frac Ad \in D_\gt.$

Let $$m = |D_\lt| = |D_\gt|$$. Then $$|D| = |D_=| + |D_\lt| + |D_\gt| = |D_=| + 2m = N$$. Note that $$|D_=|$$ has the same parity as $$N$$, and $$m = \frac{N - |D_=|}{2}$$, so

$D_= = \begin{cases} \emptyset & \text{ if } N \text{ is even} \\ \{\sqrt{A}\} & \text{ if } N \text{ is odd}. \end{cases}$ $\prod D_= = \begin{cases} 1 & \text{ if } N \text{ is even} \\ \sqrt{A} & \text{ if } N \text{ is odd}. \end{cases}$ $(\prod D_\gt) \cdot (\prod D_\lt) = \prod_{d \in D_\gt} d \cdot \frac{A}{d} = \prod_{d \in D_\gt} A = A^m$ $P = (\prod D_\gt) \cdot (\prod D_\lt) \cdot (\prod D_=) = \begin{cases} A^{\frac{N}{2}} & \text{ if } N \text{ is even} \\ A^{\frac{N - 1}{2}} \sqrt{A} = A^{\frac{N}{2}} & \text{ if } N \text{ is odd}. \end{cases}$ Therefore, $$P = A^{\frac{N}{2}}$$, so $$A = \sqrt[N]{P^2}$$.

### Credit

This is an original puzzle from cotpi.