Euler proved the identity:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\product_{p}(1-p^{-s})=1/\zeta(s) [/tex]

he made use of the fact that f(x)=x^{-s} was multiplicative..but what would have happened if f(x) were of the form:

[tex]f(x+y)=f(x)*f(y) [/tex] (1)

then if Goldbach conjecture were correct we have that :

[tex] p1+p2=Even [/tex] [tex]p3+p4+p5=Odd [/tex] for n>5 then for our function f(x):

[tex] f(p1+p2)=f(2n)=f(p1)*f(p2) [/tex]

[tex] f(p3+p4+p5)=f(2n´+1)=f(p3)*f(p4)*f(p5) [/tex]s o i think we would

have that for every function satisfying (1) we should have that:

[tex] \product_{p}(1+f(p))=\sum{f(n)}+ R [/tex]

where the sum is almost all positive integers but 0 and others that can,t be represented as a sum of primes these terms are included inside the R term.

Of ocurse this is not truly rigorous is only intuitive..:tongue2: :tongue2: :tongue2: i,m not completely sure if this all is true.

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# Curious identity?

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