Problem
NT-B2-M10-P016 Product of Divisors
#16
★★★★☆ Level 4 of 5
Prove that the product of all positive divisors of \(n\) is \(n^{\tau(n)/2}\).
Group divisors into pairs \(d\) and \(n/d\).
Every divisor \(d\) corresponds to \(\frac{n}{d}\), and the product of the pair is \(n\). If \(n\) is not a square, all divisors split into \(\frac{\tau(n)}{2}\) pairs, giving \(n^{\tau(n)/2}\). If \(n\) is a square, one divisor \(\sqrt{n}\) pairs with itself; the other pairs give \(n^{(\tau(n)-1)/2}\), and the remaining factor \(\sqrt{n}\) gives \(n^{1/2}\). The total is \(n^{\tau(n)/2}\).
This formula is often used in divisor-product problems.