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I am solving this same very question:

For any positive integer $n$, show that $\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d)$

I want to approach this question via proving the multiplicativity of the functions on both the sides, and I can take it over from there. I have easily proved that the LHS is multiplicative, however I am not able to prove the multiplicativity of the expression on the RHS. For proving the RHS to be multiplicative, I have tried proving that the expression inside the sigma is multiplicative, I think that would prove my statement but I have not been able to do so.

I also have the same problem with this question as well

Prove that $\sum \limits_{d|n}(n/d)\sigma(d) = \sum \limits_{d|n}d\tau(d)$

I want to prove that both the sides are multiplicative as well, but am unable to do so.In the above link, the questioner has already told that it is multiplicative, please help. Any help would be appreciated

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Just need these two facts to prove $f(n)=n/d \tau(d)$ is multiplicative :
1) $\tau$ is multiplicative
2) if $n = pq$ and $\gcd(p,q) = 1$, then any divisor $d$ of $n$ is of form $d_1d_2$ where $d_1 |p$ and $d_2|q$

With those two facts the proof becomes easy :

$f(pq) = \dfrac{pq}{d} \tau(d) = \dfrac{pq}{d_1d_2}\tau(d_1d_2) = \dfrac{pq}{d_1d_2}\tau(d_1)\tau(d_2) = \dfrac{p}{d_1}\tau(d_1)*\dfrac{q}{d_2}\tau(d_2)=f(p)f(q)$

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