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Recall that the Liouville function $\lambda$ and $\sigma$ are multiplicative, and the product of multiplicative functions is also multiplicative, thus $\lambda\sigma$ is multiplicative and therefore the function $\sum_{d\mid n}\lambda(d)\sigma(d)$ is multiplicative.

In the same way the function $N$ is multiplicative and therefore the function $\sum_{d\mid n}N(d) = \sum_{d\mid n}N(\frac{n}{d})$, which in particular will remain multiplicative if we only go through some divisors, so the function $\sum_{d^2\mid n} N(\frac{n}{d^2})$ is multiplicative. Therefore $\lambda(n)\sum_{d^2\mid n} N(\frac{n}{d^2}) = \lambda(n)\sum_{d^2\mid n}\frac{n}{d^2} = n\lambda(n)\sum_{d^2\mid n}\frac{1}{d^2}$ is multiplicative because it is the product of two multiplicative functions.

As both sides of the equality is a multiplicative function, it is enough to prove for the case where $n=p^k$ where $p$ is a prime. So let's start from the left side of equality.

Now let's see for cases on $k$.

Case 1: If $k$ is even $\Rightarrow k = 2l$ where $l\in\mathbb{Z}_0 ^ + $.

\begin{align*} \require{cancel} &\sigma(1)-\sigma(p)+\sigma(p^2)-\sigma(p^3)+\ldots+(-1)^{2l-1}\sigma(p^{2l-1})+(-1)^{2l}\sigma(p^{2l})\\ &= 1-(1+p)+(1+p+p^2)-(1+p+p^2+p^3)+\ldots-(1+p+p^2+p^3+\ldots+p^{2l-1})\\ &+(1+p+p^2+p^3+\ldots+p^{2l-1}+p^{2l})\\ &= 1-\cancel{(1+p)}+(\cancel{1+p}+p^2)-\cancel{(1+p+p^2+p^3)}+\ldots-\cancel{(1+p+p^2+p^3+\ldots+p^{2l-1})}\\ &+(\cancel{1+p+p^2+p^3+\ldots+p^{2l-1}}+p^{2l})\\ &= 1+p^2+p^4+\ldots+p^{2l-2}+p^{2l}\\ &=(-1)^{2l}(1+p^2+p^4+\ldots+p^{2l-2}+p^{2l})\\ &=\lambda(p^{2l})\left(\frac{p^{2l}}{p^{2l}}+\frac{p^{2l}}{p^{2l-2}}+\frac{p^{2l}}{p^{2l-4}}+\ldots+\frac{p^{2l}}{p^{2}}+\frac{p^{2l}}{p^{2(0)}}\right)\\ &=\lambda(p^{2l})\left(\frac{p^{2l}}{(p^{l})^2}+\frac{p^{2l}}{(p^{l-1})^2}+\frac{p^{2l}}{(p^{l-2})^2}+\ldots+\frac{p^{2l}}{(p)^{2}}+\frac{p^{2l}}{(p^{0})^2}\right)\\ &=\lambda(p^{2l})\sum_{i=0}^l\frac{p^{2l}}{(p^{i})^2}\\ &=\lambda(p^{2l})\sum_{d^2\mid p^{2l}}\frac{p^{2l}}{d^2}\\ &=\lambda(p^{k})\sum_{d^2\mid p^{k}}\frac{p^{k}}{d^2}\\ &=p^k\lambda(p^{k})\sum_{d^2\mid p^{k}}\frac{1}{d^2}. \end{align*}

Observation: $(-1)^{2l}=1$.

Case 2: If $k$ is odd $\Rightarrow k = 2l + 1$ where $l\in\mathbb{Z}_0^+$.

\begin{align*} \require{cancel} &\sigma(1)-\sigma(p)+\sigma(p^2)-\sigma(p^3)+\ldots+(-1)^{2l}\sigma(p^{2l})+(-1)^{2l+1}\sigma(p^{2l+1})\\ &= 1-(1+p)+(1+p+p^2)-(1+p+p^2+p^3)+\ldots+(1+p+p^2+p^3+\ldots+p^{2l})\\ &-(1+p+p^2+p^3+\ldots+p^{2l}+p^{2l+1})\\ &= 1-\cancel{(1+p)}+(\cancel{1+p}+p^2)-\cancel{(1+p+p^2+p^3)}+\ldots+(\cancel{1+p+p^2+p^3+\ldots}+p^{2l})\\ &-(1+p+p^2+p^3+\ldots+p^{2l}+p^{2l+1})\\ &= 1+p^2+p^4+\ldots+p^{2l}-(1+p+p^2+p^3+\ldots+p^{2l}+p^{2l+1})\\ &= \cancel{1}+\cancel{p^2}+\cancel{p^4}+\ldots+\cancel{p^{2l}}-(\cancel{1}+p+\cancel{p^2}+p^3+\ldots+\cancel{p^{2l}}+p^{2l+1})\\ &= -p-p^3-p^5-\ldots-p^{2l-1}-p^{2l+1}\\ &= -(p+p^3+p^5+\ldots+p^{2l-1}+p^{2l+1})\\ &=(-1)^{2l+1}(p+p^3+p^5+\ldots+p^{2l-1}+p^{2l+1})\\ &=\lambda(p^{2l+1})\left(\frac{p^{2l+1}}{p^{2l}}+\frac{p^{2l+1}}{p^{2l-2}}+\frac{p^{2l+1}}{p^{2l-4}}+\ldots+\frac{p^{2l+1}}{p^{2}}+\frac{p^{2l+1}}{p^{2(0)}}\right)\\ &=\lambda(p^{2l+1})\left(\frac{p^{2l+1}}{(p^{l})^2}+\frac{p^{2l+1}}{(p^{l-1})^2}+\frac{p^{2l+1}}{(p^{l-2})^2}+\ldots+\frac{p^{2l+1}}{(p)^{2}}+\frac{p^{2l+1}}{(p^{0})^2}\right)\\ &=\lambda(p^{2l+1})\sum_{i=0}^l\frac{p^{2l+1}}{(p^{i})^2}\\ &=\lambda(p^{2l+1})\sum_{d^2\mid p^{2l+1}}\frac{p^{2l+1}}{d^2}\\ &=\lambda(p^{k})\sum_{d^2\mid p^{k}}\frac{p^{k}}{d^2}\\ &=p^k\lambda(p^{k})\sum_{d^2\mid p^{k}}\frac{1}{d^2}. \end{align*}

Observation: $(-1)^{2l+1}=-1$.

In both cases, the same expression is reached, therefore:

\begin{align*} \sum_{d\mid p^k}\lambda(d)\sigma(d) = p^k\lambda(p^{k})\sum_{d^2\mid p^{k}}\frac{1}{d^2}. \end{align*}

From what was said at the beginning, since both sides of the equation were multiplicative, it was enough to test for the power of a prime, so we have tested equality for all $n$.

Note: I hope that this time Xander Henderson, Professor Vector, Peter and amWhy do not close my publication because apparently they cannot ask for help (I clarify help, not solutions) and I leave my solution in case someone wants to take it as an example of how to address this type of problem, already that if someone asks only for some ideas, they may close the question.

BML
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1 Answers1

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Probably correct, anyway you got the idea.

I think it is ok to show how I deal with it

$$\sum_{n\ge 1} \sigma(n)n^{-s}=\sum_{n\ge 1} (\sum_{d|n} d)n^{-s} = \prod_p \frac1{(1-p^{-s})(1-p^{1-s})}$$ $$\sum_{n\ge 1} \lambda(n)\sigma(n)n^{-s} = \prod_p \frac1{(1-\lambda(p)p^{-s})(1-\lambda(p) p^{1-s})}= \prod_p \frac1{(1+p^{-s})(1+p^{1-s})}$$ $$\sum_{n\ge 1}(\sum_{d|n} \lambda(d)\sigma(d))n^{-s} =\prod_p \frac1{(1-p^{-s})(1+p^{-s})(1+p^{1-s})}=\prod_p \frac1{(1-p^{-2s})(1+p^{1-s})}$$

$$\sum_{n\ge 1}(\sum_{d^2|n} d^{-2})n^{-s} = \zeta(s)\sum_{d\ge 1}d^{-2s-2}= \zeta(s)\zeta(2s+2)=\prod_p \frac1{(1-p^{-s})(1-p^{-2s-2})}$$ $$\sum_{n\ge 1}n(\sum_{d^2|n} d^{-2})n^{-s} = \prod_p \frac1{(1-p^{1-s})(1-p^{-2s})}$$

$$\sum_{n\ge 1}n\lambda(n)(\sum_{d^2|n} d^{-2})n^{-s} = \prod_p \frac1{(1-\lambda(p)p^{1-s})(1-\lambda(p^2)p^{-2s})}= \prod_p \frac1{(1+p^{1-s})(1-p^{-2s})}$$

Whence $$\sum_{d|n} \lambda(d)\sigma(d)=n\lambda(n)(\sum_{d^2|n} d^{-2})$$

reuns
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