s=s... again?

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we remember that n=1Ch(n)ns=2ζ(2s)+ζ(s)\sum_{n=1}^\infty\dfrac{Ch(n)}{n^s}=-2\zeta(2s)+\zeta(s) consider Cs(n)=(Ch1)(n)=dnCh(d)Cs(n)=(Ch*1)(n)=\sum_{d|n}Ch(d) so, if d is a perfect square it is -1, 1 otherwise. if there are k perfect square divisors, there are τ(n)k\tau(n)-k non-square divisor. add them and it is τ(n)2k\tau(n)-2k.

let k be a function of n. It is multiplicative. So we can compute for prime powers k(pa)=d2pad0=10+(p2)0+(p4)0+....(p2a2)0=a2+1k(p^a)=\sum_{d^2|p^a} d^{0}=1^0+(p^2)^0+(p^4)^0+.... \left(p^{2\lfloor \dfrac{a}{2}\rfloor}\right)^0=\lfloor \dfrac{a}{2}\rfloor+1 we have n=1Ch(n)nsn=11ns=n=1τ(n)ns2n=1k(n)ns\sum_{n=1}^\infty\dfrac{Ch(n)}{n^s}\sum_{n=1}^\infty\dfrac{1}{n^s}=\sum_{n=1}^\infty\dfrac{\tau(n)}{n^s}-2\sum_{n=1}^\infty\dfrac{k(n)}{n^s} 2ζ(s)ζ(2s)+ζ2(s)=ζ2(s)2n=1k(n)ns-2\zeta(s)\zeta(2s)+\zeta^2(s)=\zeta^2(s)-2\sum_{n=1}^\infty\dfrac{k(n)}{n^s} n=1k(n)ns=ζ(s)ζ(2s)\sum_{n=1}^\infty\dfrac{k(n)}{n^s}=\zeta(s)\zeta(2s) Lets try to compute the summation using the fact it is multpicative p=primea0k(pa)pas=p=primea0a2+1pas=p=primep3s(ps1)2(ps+1)=p1(1ps)21ps1p2s=ζ2(s)ζ(2s)ζ(s)=ζ(s)ζ(2s)\prod_{p=prime} \sum_{a≥0}\dfrac{k(p^a)}{p^{as}}=\prod_{p=prime} \sum_{a≥0}\dfrac{\lfloor \dfrac{a}{2}\rfloor+1}{p^{as}}=\prod_{p=prime} \dfrac{p^{3s}}{(p^s-1)^2(p^s+1)}\\=\prod_p \dfrac{1}{(1-p^{-s})^2}\dfrac{1-p^ {-s}}{1-p^{-2s}}=\zeta^2(s)\dfrac{\zeta(2s)}{\zeta(s)}=\zeta(s)\zeta(2s) so we get s=s\boxed{\large{s=s}} again!!!

Note by Aareyan Manzoor
5 years, 5 months ago

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The combo part was not found by me but @Andrew Ellinor so he gets the credit for that.

He was not sure wether or not he was correct but this proves that k(n)=i=1ω(n)(ki2+1)k(n)=\prod_{i=1}^{\omega(n)} (\lfloor \dfrac{k_i}{2}\rfloor +1) if n=p1k1p2k2.....pω(n)kω(n)n=p_1^{k_1}p_2^{k_2}.....p_{\omega(n)}^{k_{\omega(n)}}

Aareyan Manzoor - 5 years, 5 months ago

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Well he is correct.

Julian Poon - 5 years, 5 months ago

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