Are there any solutions to the equation $\zeta(a_0+bi)=\zeta(a_1+bi)$ where $a_0\neq a_1\neq0$ and $b\neq0$? If so, how would one find these solutions? Without the restrictions set up, you could easily create trivial solutions such as $(a_0,a_1,b)=(a,a,b)$ or $ (a_0,a_1,b)=(-2m, -2n, b)$ where $m,n$ are whole numbers.
I would also like to ask if there are any algorithms besides brute force algorithms to find solutions to this.
I can't find a rigorous argument for $\clubsuit$ but I think this is the right track :
Let $$F_a(s) = \frac{2^s}{1-2^{-a}}(\zeta(s)-\zeta(s+a)) = 1+ \sum_{n=3}^\infty (n/2)^{-s}\frac{1-n^{-a}}{1-2^{-a}}$$ Since $F_a(s) \to 1$ uniformly as $Re(s) \to \infty$, no problem for writing that for $Re(s)$ large enough : $$\frac{1}{F_a(s)} = \frac{1}{1-(1-F_a(s))} = \sum_{k=0}^\infty (1-F_a(s))^k = 1+\sum_{k=1}^\infty (-1)^k\sum_{n=3}^\infty c_a(n,k)(n/2^k)^{-s}$$ (where $c_a(n,k) = \sum_{n = m_1\ldots m_k}\prod_{j=1}^k\frac{1-m_j^{-a}}{1-2^{-a}}$)
It is kind of obvious ($\clubsuit$ why ?) that this generalized Dirichlet series has an abscissa of convergence $\sigma_a \ge 1/2$, and since it is clearly meromorphic, it should mean there is at least one $z, Re(z) =\sigma_a$ such that $F_a(z) =0$ i.e. $\zeta(z) = \zeta(z+a)$