Alfonso De Miguel Bueno Ademiguelbueno@gmail.com April 21, 2020. Madrid, Spain Hidden Asymmetries in Riemann Zeta Function to Refute the Riemann Hypothesis I'm going to try to summarize here a previous article about prime numbers to especially bold the case where an asymmetry between the complex conjugates non-trivial zeros of the Riemann Zeta function takes place, which could represent a counter example of the Riemann Hypothesis. The Riemann Hypothesis states that all non-trivial zeros of the Riemann Z function have a real part one half. Let's verify it by firstly creating separate functions for every prime number, and then overlapping them on the function of the prime 3 to see if they create interferences on a cycle of the function 3. To build such functions I drew two columns for every prime, a column for odd numbers and a column for even numbers. The cycle of each function is given by adding 2 times the value of the prime number to that prime itself, for example for the case of 3, we add 3+3+3 getting a cycle whose length goes from 3 to 9. Then, let's trace a dotted line from the valley of each function different of 3 to project them on the function 3. When those projected lines overlap the valley of the cycles of the function 3 no interference will happen. But when thy overlap an amplitude of the function 3 an interference is going to take place on the affected - number of the function 3. The first cycle of each function different than 3 won't interfere, and without computing such first cycle, no interference will take place every 3 cycles, because their valleys will overlap, those cycles will be divisible by 3. Let's see as an example the non-prime number 15 on function 3. The valley of the first cycle of function 5 overlaps the valley of the second cycle of function 3. That -15 point of function 3 will be a trivial zero. It's a zero in the sense of the Riemann Z function because we reduce the value of the even +10 to ceros from function 5 to function 3, being located its ceros at the point where -15 is located. Then, let's check what happens with -25 on function 3. We see that the valley of the second cycle of function 5 interferes with the amplitude of the fourth cycle of the function 3. In that case, we have reduced to zeros the even +20 from function five to function 3, reaching that zero value at the valley where the point -25 of function 3 is placed. That point will be a non-trivial zero.