Toolkit 35

Product of Positive Divisors

n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}

t=(\alpha_1+1)(\alpha_2+1)\cdots(\alpha_k+1)

\text{Product of positive divisors of }n:\;n^{t/2}

Proof

Let $t$ be the number of positive divisors of $n$ , and let the positive divisors of $n$ be

d_1, d_2, \ldots, d_t.

If $d$ is a positive divisor of $n$ , then $n/d$ is also a positive divisor of $n$ . Thus, the divisors can be paired so that each pair has product $n$ :

d_1d_t = n, \qquad d_2d_{t-1} = n, \qquad \ldots

Let

D = d_1d_2\cdots d_t

be the product of all positive divisors. Since the map $d \mapsto n/d$ permutes the positive divisors, we also have

D = \frac{n}{d_1}\cdot\frac{n}{d_2}\cdots\frac{n}{d_t} = \frac{n^t}{D}.

Therefore,

D^2 = n^t.

Since $D > 0$ ,

D = n^{t/2}. \quad\square

This argument also includes the case in which $n$ is a perfect square: the divisor $\sqrt{n}$ is paired with itself.