这是一份southampton南安普敦大学MATH3078W1-011作业代写的成功案例
If $p$ is a prime and $k \geq 0$ we have
$$
\sigma\left(p^{k}\right)=\frac{p^{k+1}-1}{p-1} .
$$
Since $p$ is prime, the divisors of $p^{k}$ are $1, p, p^{2}, \ldots, p^{k}$. Hence
$$
\sigma\left(p^{k}\right)=1+p+p^{2}+\cdots+p^{k}=\frac{p^{k+1}-1}{p-1}
$$
as desired.
Let $n=p_{1}^{e_{1}} p_{2}^{e_{2}} \cdots p_{r}^{e_{r}}$. Our proof is by induction on $r$. If $r=1, n=p_{1}^{e_{1}}$ and the result follows. Suppose the result is true when $1 \leq r \leq k$. Consider now the case $r=k+1$. That is, let
$$
n=p_{1}^{e_{1}} \cdots p_{k}^{e_{k}} p_{k+1}^{e_{k+1}}
$$
MATH3078W1-01 COURSE NOTES ：
$$
\sigma(n)=\sigma\left(2^{k}\right) \sigma(q)=\left(2^{k+1}-1\right) \sigma(q)
$$
So we have
$$
2^{k+1} q=2 n=\sigma(n)=\left(2^{k+1}-1\right) \sigma(q),
$$
hence
() $2^{k+1} q=\left(2^{k+1}-1\right) \sigma(q) .$ Now $\sigma^{}(q)=\sigma(q)-q$, so $$ \sigma(q)=\sigma^{}(q)+q . $$ Putting this in () we get
$$
2^{k+1} q=\left(2^{k+1}-1\right)\left(\sigma^{*}(q)+q\right)
$$