My solution depends on the following hypothesis. Can anyone prove this?

~~Sorry, the image doesn't work for some reason. \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{c} d(ijk) = \sum_{gcd(i,j)=gcd(j,k)=gcd(k,i)=1} [\frac{a}{i}][\frac{b}{j}][\frac{c}{k}] Now it works.~~

Sorry, please type "http://rng.gozaru.jp/cf.png" directly to the address bar.

Finally proved: "http://rng.gozaru.jp/b.pdf"

Image is not shown because of 403 error

I have thought about it for a long time...still can't understand it...It's like magic...

I think it can even be extended to 4 or larger number.

Sure. Proof will be the same

Does this couclusion can still be hold in more general form ? .. .

Can we prove it by Möbius_inversion_formula ? .

Your proof one could easily reformulate without telescopy. We have sum of

d(ijk) -- number of triplesx,y,zpairwise coprime divisors ofi,j,krespectively. Consider how many times we count any triple of coprime integersx,y,z. Obviously it is a product of their multipliers numbers, it equalsThanks for the neat and instructive proof.