I proved Riemann Hypothesis: let f(n) be the function of the first n primes, so that f(n)=1 if and only if n is prime. Let F(x) be the sum of all the f(n), and then prove F(x)=0. If F(x)=0, then Riemann Hypothesis is true. If F(x)>0, then there is an R that satisfies 1<R<x so that for n=R+1, R does not divide n+1. Now let g(n) be the product of all the numbers R, satisfying 1<R<x, but for n=R+1. Let F(x) = f(R)+f(R+1)+...+f(R+n), and then the g(n) must satisfy F(x)=0, which contradicts the definition of f(n) for this R. Thus, if F(x)>0, then Riemann Hypothesis is false. Q.E.D.