Sum of ϕ of divisors
Dependencies:
Let $\phi$ be Euler's totient function (which means $|\mathbb{Z}_n^*| = \phi(n)$). Then,
\[ n = \sum_{d \mid n} \phi(d) \]
Proof
$\mathbb{Z}_n$ can be partitioned into $\left\{ d \mathbb{Z}_{\frac{n}{d}}^* : d \mid n\right\}$.
Therefore, \[ n = |\mathbb{Z}_n| = \sum_{d \mid n} \left|d \mathbb{Z}_{\frac{n}{d}}^*\right| = \sum_{d \mid n} \phi\left(\frac{n}{d}\right) = \sum_{d \mid n} \phi(d) \]
Dependency for: None
Info:
- Depth: 5
- Number of transitive dependencies: 6