- #1

Math100

- 771

- 219

- Homework Statement
- By considering ## u\cdot u\cdot N\cdot N ##, prove that ## \sum_{d\mid n}\sigma_{1}(d)\sigma_{1}(\frac{n}{d})=\sum_{d\mid n}d\sigma_{0}(d)\sigma_{0}(\frac{n}{d}) ##.

- Relevant Equations
- None.

Proof:

Let ## n\in\mathbb{N} ##.

Then ## \sigma_{\alpha}(n)=(u\cdot N^{\alpha})(n) ## for ## \alpha\neq 0 ## and ## \sigma_{0}(n)=(u\cdot u)(n) ##

such that ## u(n)=1, N(n)=n ## for all ## n ##.

By considering ## u\cdot u\cdot N\cdot N ##, we have that

\begin{align*}

&\sigma_{1}\cdot \sigma_{1}=(u\cdot N)\cdot (u\cdot N)\\

&=(N\cdot u)\cdot (u\cdot N)\\

&=N\cdot (u\cdot u)\cdot N\\

&=N\cdot \sigma_{0}\cdot N\\

&=\sigma_{0}\cdot (N\cdot N).\\

\end{align*}

Observe that

\begin{align*}

&(\sigma_{1}\cdot \sigma_{1})(n)=[\sigma_{0}\cdot (N\cdot N)](n)\\

&=\sum_{d\mid n}\sigma_{0}(\frac{n}{d})(N\cdot N)(d)\\

&=\sum_{d\mid n}\sigma_{0}(\frac{n}{d})\sum_{d_{1}\mid d}N(d_{1})N(\frac{d}{d_{1}})\\

&=\sum_{d\mid n}\sigma_{0}(\frac{n}{d})\sum_{d_{1}\mid d}d_{1}(\frac{d}{d_{1}})\\

&=\sum_{d\mid n}\sigma_{0}(\frac{n}{d})\sum_{d_{1}\mid d}d\\

&=\sum_{d\mid n}\sigma_{0}(\frac{n}{d})d\sum_{d_{1}\mid d}1\\

&=\sum_{d\mid n}\sigma_{0}(\frac{n}{d})d\sum_{d_{1}\mid d}u(d_{1})u(\frac{d}{d_{1}})\\

&=\sum_{d\mid n}\sigma_{0}(\frac{n}{d})d[(u\cdot u)(d)]\\

&=\sum_{d\mid n}d\sigma_{0}(d)\sigma_{0}(\frac{n}{d}).\\

\end{align*}

Since ## (\sigma_{1}\cdot \sigma_{1})(n)=\sum_{d\mid n}\sigma_{1}(d)\sigma_{1}(\frac{n}{d}) ##,

it follows that ## \sum_{d\mid n}\sigma_{1}(d)\sigma_{1}(\frac{n}{d})=\sum_{d\mid n}d\sigma_{0}(d)\sigma_{0}(\frac{n}{d}) ##.

Therefore, ## \sum_{d\mid n}\sigma_{1}(d)\sigma_{1}(\frac{n}{d})=\sum_{d\mid n}d\sigma_{0}(d)\sigma_{0}(\frac{n}{d}) ##.

Let ## n\in\mathbb{N} ##.

Then ## \sigma_{\alpha}(n)=(u\cdot N^{\alpha})(n) ## for ## \alpha\neq 0 ## and ## \sigma_{0}(n)=(u\cdot u)(n) ##

such that ## u(n)=1, N(n)=n ## for all ## n ##.

By considering ## u\cdot u\cdot N\cdot N ##, we have that

\begin{align*}

&\sigma_{1}\cdot \sigma_{1}=(u\cdot N)\cdot (u\cdot N)\\

&=(N\cdot u)\cdot (u\cdot N)\\

&=N\cdot (u\cdot u)\cdot N\\

&=N\cdot \sigma_{0}\cdot N\\

&=\sigma_{0}\cdot (N\cdot N).\\

\end{align*}

Observe that

\begin{align*}

&(\sigma_{1}\cdot \sigma_{1})(n)=[\sigma_{0}\cdot (N\cdot N)](n)\\

&=\sum_{d\mid n}\sigma_{0}(\frac{n}{d})(N\cdot N)(d)\\

&=\sum_{d\mid n}\sigma_{0}(\frac{n}{d})\sum_{d_{1}\mid d}N(d_{1})N(\frac{d}{d_{1}})\\

&=\sum_{d\mid n}\sigma_{0}(\frac{n}{d})\sum_{d_{1}\mid d}d_{1}(\frac{d}{d_{1}})\\

&=\sum_{d\mid n}\sigma_{0}(\frac{n}{d})\sum_{d_{1}\mid d}d\\

&=\sum_{d\mid n}\sigma_{0}(\frac{n}{d})d\sum_{d_{1}\mid d}1\\

&=\sum_{d\mid n}\sigma_{0}(\frac{n}{d})d\sum_{d_{1}\mid d}u(d_{1})u(\frac{d}{d_{1}})\\

&=\sum_{d\mid n}\sigma_{0}(\frac{n}{d})d[(u\cdot u)(d)]\\

&=\sum_{d\mid n}d\sigma_{0}(d)\sigma_{0}(\frac{n}{d}).\\

\end{align*}

Since ## (\sigma_{1}\cdot \sigma_{1})(n)=\sum_{d\mid n}\sigma_{1}(d)\sigma_{1}(\frac{n}{d}) ##,

it follows that ## \sum_{d\mid n}\sigma_{1}(d)\sigma_{1}(\frac{n}{d})=\sum_{d\mid n}d\sigma_{0}(d)\sigma_{0}(\frac{n}{d}) ##.

Therefore, ## \sum_{d\mid n}\sigma_{1}(d)\sigma_{1}(\frac{n}{d})=\sum_{d\mid n}d\sigma_{0}(d)\sigma_{0}(\frac{n}{d}) ##.