NT-B2-M10-P014 When \(\varphi(n)\mid n\)

Look at \(\frac{n}{\varphi(n)}=\prod_{p\mid n}\frac{p}{p-1}\) and at the number of odd prime divisors.

For \(\frac{n}{\varphi(n)}\) to be an integer, the product \(\prod_{p\mid n}\frac{p}{p-1}\) must be integral. Every odd prime \(p\) contributes an even factor \(p-1\) in the denominator. In the numerator there can be only one factor \(2\), if \(2\mid n\). Hence there is at most one odd prime divisor. If such a prime \(p\) occurs, we need \(p-1\mid 2\), so \(p=3\), and \(2\mid n\) is necessary. Therefore the solutions are \(n=1\), \(n=2^a\) for \(a\ge 1\), and \(n=2^a3^b\) for \(a,b\ge 1\). All are easily checked.

An olympiad-style problem restricting the set of prime divisors.