a quite evident for rigorous proof each root was the solution of any non-trivial zeros are on the...

proof: let any complex number s=x+i*tn,then zeta(s)=zeta(s)/2^s+f(s)/(3^s -1) the zeta(1-s)=f(1-s)/(3^(1-s)-1)(1-1/2^(1-s)),because of for any sufficiently large known prime positive odd number the new complex function  equation f(s)=f(1-s) always have the same solution root,that was tn(s)=tn(1-s);so x+tn(s)^2/Re(s)=(1-x)+(tn(1-s))^2/Re(1-s),even any value of tn tends to infinity biggest real number, that is only and if only Re(s)=1/2,all non trivial zeros are on the line x=1/2,because of the quite evident correctly focuses on the original basic case of zeta,the more deeper information will led to many real progress and look into the difficult proposition future.more sophisticated results are all larger enough odd prime behavior were countable real numbers elements sets,the numerical and theoretical confirmation of Riemann Hypothesis is again very striking.completely proof the Riemann hypothesis proposition.more precisely each Im(s)=t(p)value also countable real numbers.