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Let: $$\rho=\frac{1}{1-\frac{A}{B}}+\frac{1}{n}+s$$

See: https://mathoverflow.net/a/439430/25104

Let: $$\rho=\frac{1}{1-\frac{A}{B}}+\frac{1}{n}+s$$

$$\tag{2}$$ $A=\frac{B \cdot n \cdot \rho -B \cdot n \cdot s+B (-(n+1))}{n \cdot \rho-n \cdot s-1}$ $\land \left(\left(\Im(\rho)<\Im(s)\land \Re(\rho)=\frac{1}{2}\land (\Re(B)<0\lor (\Re(B)=0\land (\Im(B)<0\lor \Im(B)>0))\lor \Re(B)>0)\right)\lor \left(\Im(\rho)=\Im(s)\land \left(\left(\Re(s)<\frac{n-2}{2 n}\land \Re(\rho)=\frac{1}{2}\land (\Re(B)<0\lor (\Re(B)=0\land (\Im(B)<0\lor \Im(B)>0))\lor \Re(B)>0)\right)\lor \left(\Re(s)>\frac{n-2}{2 n}\land \Re(\rho)=\frac{1}{2}\land (\Re(B)<0\lor (\Re(B)=0\land (\Im(B)<0\lor \Im(B)>0))\lor \Re(B)>0)\right)\right)\right)\lor \left(\Im(\rho)>\Im(s)\land \Re(\text{rho})=\frac{1}{2}\land (\Re(B)<0\lor (\Re(B)=0\land (\Im(B)<0\lor \Im(B)>0))\lor \Re(B)>0)\right)\right)\land 0\leq \Re(\rho)\leq 1$

$$\tag{2}$$ $A=\frac{B \cdot n \cdot \rho -B \cdot n \cdot s+B (-(n+1))}{n \cdot \rho-n \cdot s-1}$ $\land \left(\left(\Im(\rho)<\Im(s)\land \Re(\rho)=\frac{1}{2}\land (\Re(B)<0\lor (\Re(B)=0\land (\Im(B)<0\lor \Im(B)>0))\lor \Re(B)>0)\right)\lor \left(\Im(\rho)=\Im(s)\land \left(\left(\Re(s)<\frac{n-2}{2 n}\land \Re(\rho)=\frac{1}{2}\land (\Re(B)<0\lor (\Re(B)=0\land (\Im(B)<0\lor \Im(B)>0))\lor \Re(B)>0)\right)\lor \left(\Re(s)>\frac{n-2}{2 n}\land \Re(\rho)=\frac{1}{2}\land (\Re(B)<0\lor (\Re(B)=0\land (\Im(B)<0\lor \Im(B)>0))\lor \Re(B)>0)\right)\right)\right)\lor \left(\Im(\rho)>\Im(s)\land \Re(\rho)=\frac{1}{2}\land (\Re(B)<0\lor (\Re(B)=0\land (\Im(B)<0\lor \Im(B)>0))\lor \Re(B)>0)\right)\right)\land 0\leq \Re(\rho)\leq 1$

So in other words there are choices of $A$, $B$ and the seed point $s$ where $\Re(\rho) = 1/2$, but the conditions above must be satisfied in order for the following equation:

(* Mathematica start *)
Clear[A, B, n, k, s];
n = 20;
s = 0;
A = Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/Zeta[s + k/n], {k, 1, n}];
B = Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/Zeta[s + k/n + 1/n], {k, 
    1, n}];
"The first trivial zero with the zero of the conjugated limit shifted \
by 1 and with opposite sign:"
Conjugate[-s - 1/n + 1/(1 - B/A)];
N[%, n]
 s + 1/n + 1/(1 - A/B);
N[%, n]

Clear[A, B, n, k, s];
n = 20;
s = (1/3 + 14*I);
A = Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/Zeta[s + k/n], {k, 1, n}];
B = Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/Zeta[s + k/n + 1/n], {k, 
    1, n}];
"The first non-trivial zero:"
Conjugate[-s - 1/n + 1/(1 - B/A)];
N[%, n]
 s + 1/n + 1/(1 - A/B);
N[%, n]

"The conditions to be proven for the case of the Riemann zeta function:"
Clear[s, A, B, n, z, k];
n = 19;
Reduce[z == s + 1/n + 1/(1 - A/B) && 
  s + 1/n + 1/(1 - A/B) == Conjugate[-s - 1/n + 1/(1 - B/A)] && 
  B != 0 && Re[z] >= 0 && Re[z] <= 1, Re[z]]
(*end*)

So in other words there are choices of $A$, $B$ and the starting point $s$ where $\Re(\rho) = 1/2$, but the conditions above must be satisfied in order for the following equation:

(* Mathematica start *)
Clear[A, B, n, k, s];
n = 20;
s = 0;
A = Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/Zeta[s + k/n], {k, 1, n}];
B = Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/Zeta[s + k/n + 1/n], {k, 
    1, n}];
"The first trivial zero with the zero of the conjugated limit shifted \
by 1 and with opposite sign:"
Conjugate[-s - 1/n + 1/(1 - B/A)];
N[%, n]
 s + 1/n + 1/(1 - A/B);
N[%, n]

Clear[A, B, n, k, s];
n = 20;
s = (1/3 + 14*I);
A = Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/Zeta[s + k/n], {k, 1, n}];
B = Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/Zeta[s + k/n + 1/n], {k, 
    1, n}];
"The first non-trivial zero:"
Conjugate[-s - 1/n + 1/(1 - B/A)];
N[%, n]
 s + 1/n + 1/(1 - A/B);
N[%, n]

"The conditions to be proven for the case of the Riemann zeta function:"
Clear[s, A, B, n, z, k];
n = 19;
Reduce[rho == s + 1/n + 1/(1 - A/B) && 
  s + 1/n + 1/(1 - A/B) == Conjugate[-s - 1/n + 1/(1 - B/A)] && 
  B != 0 && Re[rho] >= 0 && Re[rho] <= 1, Re[rho]]
(*end*)