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Sorry, I let mind to wander around in that post, as I wanted to get a result also for the case when x and y are not coprimes (it was rather easy to get it for when they are coprimes, so I wanted to go a bit beyond). Let me try to organize it. That factorization in prime factors was really superfluous.This is very difficult to understand. What is ddd? You only have s(xy)−s(x)s(y)=M and that x and y are coprime. That's where you start. Everything else has to be explained or defined. And then come uuu and vvv out of the blue. And how come the pairing zp=xp+yp?zp=xp+yp?z_p=x_p+y_p\,?

I'm completely confused.

##d|(x*y)##, d is a divisor of x*y

##u|x##, ##v|y##, u is a divisor of x, v is a divisor of y

from the previous results we know that every

##d=u*v## for some ##u|x##, ##v|y##

in general multiple ##u,v## satisfy ##d=u*v##, so define

N(d,x,y) = number of ##u,v## pairs that satisfy ##d=u*v, u|x, v|y##

##s(x) = \sum u## for all ##u|x##

##s(y) = \sum v## for all ##v|y##

##s(x)*s(y) = \sum \sum u*v## for all ##u|x, v|y##

##s(x)*s(y) = \sum N(d,x,y)*d## for all distinct d such that ##d=u*v, u|x, v|y##

##s(x)*s(y) = \sum d + \sum (N(d,x,y)-1)*d## for all distinct d such that ##d=u*v, u|x, v|y##

call

##S = \sum d ## for all distinct d such that ##d=u*v, u|x, v|y##

as the sum occurs over all possible u,v pairs, that will include all possible d such that ##d|(x*y)##

##S= \sum d ## for all distinct d such that ##d|(x*y)##

by definition of s(),

##S = s(x*y)##

##s(x)*s(y) = s(x*y) - \sum (N(d,x,y)-1)*d## for all distinct d such that ##d=u*v, u|x, v|y##

##s(x)*s(y) = s(x*y) - M##

##M = \sum (N(d,x,y)-1)*d##

if x and y are co-primes then either ##d=u*1## or ##d=1*v##, therefore N(d,x,y)=1 and M=0.

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