試貼數論課的作業

Show that for any positive integer n, \(\sigma (n)=\sum_{e|n} \phi (e)d(\frac{n}{e})\)

Proof:
Let \(n=p^\alpha\), p prime. Then
$$\sum_{e|n} \phi(e)d(\frac{n}{e})$$
$$=\sum_{i=0}^{\alpha}\phi(p^i)d(p^{\alpha-i})$$
$$=\phi(1)d(p^\alpha)+\phi(p)d(p^{\alpha-1})+\cdots +\phi(p^\alpha)d(1)$$
$$=(\alpha+1)+(p-1)\alpha+p(p-1)(\alpha-1)+p^{\alpha-2}(p-1)2+p^{\alpha-1}(p-1)$$
$$=(\alpha+1)+\alpha p+(\alpha-1)p^2+(\alpha-2)p^3+\cdots +2p^{\alpha-1}+p^\alpha$$
$$-\alpha -(\alpha-1)p-(\alpha-2)p^2-(\alpha-3)p^3-\cdots -p^{\alpha-1}$$
$$=1+p+p^2+\cdots +p^\alpha=\frac{p^{\alpha+1}-1}{p-1}=\sigma(p^\alpha)=\sigma(n)$$
But \(\phi , d\) and \(\sigma\) are multiplicative. So \(\phi(n)=\sum_{e|n}\phi(e)d(\frac{n}{e}), \forall n \in \mathbb{N}\)