NT-B2-M10-P006 When \(\varphi(n)=n/2\)

Use \(\frac{n}{\varphi(n)}=\prod_{p\mid n}\frac{p}{p-1}\).

The condition gives \(\frac{n}{\varphi(n)}=2\). If \(n\) has an odd prime divisor \(p\), then the product contains \(\frac{p}{p-1}\), and even with the possible factor \(2\) from the prime \(2\), the product cannot be exactly \(2\). Thus the only prime divisor of \(n\) is \(2\). Hence \(n=2^a\), \(a\ge 1\). All such numbers work since \(\varphi(2^a)=2^{a-1}\).

A classic problem on the ratio \(n/\varphi(n)\).