- #1

Math100

- 779

- 220

- Homework Statement
- Prove that ## \sum_{d\mid n}\sigma_{2}(d)\sigma_{2}(\frac{n}{d})=\sum_{d\mid n}d^{2}\sigma_{0}(d)\sigma_{0}(\frac{n}{d}) ##, and verify this result when ## n=10 ##.

- Relevant Equations
- None.

Proof:

Observe that

\begin{align*}

\sum_{d\mid n}\sigma_{2}(d)\sigma_{2}(\frac{n}{d})&=(N^{2}\ast u)\ast (N^{2}\ast u) \text{ since } \sigma_{2}(n)=\sum_{d\mid n}d^{2}\cdot 1=N^{2}\ast u)\\

&=(u\ast N^{2})\ast (N^{2}\ast u)\\

&=u\ast (N^{2}\ast N^{2})\ast u\\

&=u\ast (N^{2}\sigma_{0})\ast u \text{ since } N^{2}\ast N^{2}=\sum_{d\mid n}d^{2}(\frac{n^{2}}{d^{2}})=n^{2}\sum_{d\mid n}1=N^{2}\sigma_{0} \\

&=N^{2}\sigma_{0}\ast (u\ast u) \text{ since } u\ast u=\sigma_{0} \\

&=N^{2}\sigma_{0}\ast \sigma_{0}\\

&=\sum_{d\mid n}d^{2}\sigma_{0}(d)\sigma_{0}(\frac{n}{d}).\\

\end{align*}

Therefore, ## \sum_{d\mid n}\sigma_{2}(d)\sigma_{2}(\frac{n}{d})=\sum_{d\mid n}d^{2}\sigma_{0}(d)\sigma_{0}(\frac{n}{d}) ##.

Now we will verify this result when ## n=10 ##.

Consider the divisors ## d\in\left \{ 1, 2, 5, 10 \right \} ##.

Observe that

\begin{align*}

\sum_{d\mid 10}\sigma_{2}(d)\sigma_{2}(\frac{10}{d})&\stackrel{?}{=}\sum_{d\mid 10}d^{2}\sigma_{0}(d)\sigma_{0}(\frac{10}{d})

\end{align*}

\begin{align*}

&\sigma_{2}(1)\sigma_{2}(\frac{10}{1})+\sigma_{2}(2)\sigma_{2}(\frac{10}{2})+\sigma_{2}(5)\sigma_{2}(\frac{10}{5})+\sigma_{2}(10)\sigma_{2}(\frac{10}{10})\\

&\stackrel{?}{=}1^{2}\sigma_{0}(1)\sigma_{0}(\frac{10}{1})+2^{2}\sigma_{0}(2)\sigma_{0}(\frac{10}{2})+5^{2}\sigma_{0}(5)\sigma_{0}(\frac{10}{5})+10^{2}\sigma_{0}(10)\sigma_{0}(\frac{10}{10}).

\end{align*}

Observe that

\begin{align*}

\sum_{d\mid n}\sigma_{2}(d)\sigma_{2}(\frac{n}{d})&=(N^{2}\ast u)\ast (N^{2}\ast u) \text{ since } \sigma_{2}(n)=\sum_{d\mid n}d^{2}\cdot 1=N^{2}\ast u)\\

&=(u\ast N^{2})\ast (N^{2}\ast u)\\

&=u\ast (N^{2}\ast N^{2})\ast u\\

&=u\ast (N^{2}\sigma_{0})\ast u \text{ since } N^{2}\ast N^{2}=\sum_{d\mid n}d^{2}(\frac{n^{2}}{d^{2}})=n^{2}\sum_{d\mid n}1=N^{2}\sigma_{0} \\

&=N^{2}\sigma_{0}\ast (u\ast u) \text{ since } u\ast u=\sigma_{0} \\

&=N^{2}\sigma_{0}\ast \sigma_{0}\\

&=\sum_{d\mid n}d^{2}\sigma_{0}(d)\sigma_{0}(\frac{n}{d}).\\

\end{align*}

Therefore, ## \sum_{d\mid n}\sigma_{2}(d)\sigma_{2}(\frac{n}{d})=\sum_{d\mid n}d^{2}\sigma_{0}(d)\sigma_{0}(\frac{n}{d}) ##.

Now we will verify this result when ## n=10 ##.

Consider the divisors ## d\in\left \{ 1, 2, 5, 10 \right \} ##.

Observe that

\begin{align*}

\sum_{d\mid 10}\sigma_{2}(d)\sigma_{2}(\frac{10}{d})&\stackrel{?}{=}\sum_{d\mid 10}d^{2}\sigma_{0}(d)\sigma_{0}(\frac{10}{d})

\end{align*}

\begin{align*}

&\sigma_{2}(1)\sigma_{2}(\frac{10}{1})+\sigma_{2}(2)\sigma_{2}(\frac{10}{2})+\sigma_{2}(5)\sigma_{2}(\frac{10}{5})+\sigma_{2}(10)\sigma_{2}(\frac{10}{10})\\

&\stackrel{?}{=}1^{2}\sigma_{0}(1)\sigma_{0}(\frac{10}{1})+2^{2}\sigma_{0}(2)\sigma_{0}(\frac{10}{2})+5^{2}\sigma_{0}(5)\sigma_{0}(\frac{10}{5})+10^{2}\sigma_{0}(10)\sigma_{0}(\frac{10}{10}).

\end{align*}

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