Question: Let $A$ be a symmetric operator on the Hilbert space $\mathcal{H}$. Assume that $A$ is not self-adjoint but $\operatorname{Ran}(A-i) = \mathcal{H}$ or $\operatorname{Ran}(A+i) = \mathcal{H}$. Prove that $A$ has no self-adjoint extension.
My attempt: I know that I have to somehow use the fact that a self-adjoint operator $\tilde{A}$ has the property $\operatorname{Ran}(\tilde{A}-i) = \mathcal{H}$ and $\operatorname{Ran}(\tilde{A}+i) = \mathcal{H}$. So I assumed that it has a self-adjoint extension and tried to induce any contradiction. However, I couldn't find the connections and didn't find a way to proceed.
Note: I found something about the Deficiency Indices which seems to be the direct path to this exercise but this is not covered in class. I think this exercise should be very simple but I just can't find a way forward.
For concreteness, suppose that $\mathrm{Ran}(A+\mathrm{i})= H$. Let $\widetilde A$ be a self-adjoint extension of $A$. Since $A$ is not self-adjoint, the domain of $\widetilde A$ strictly contains that of $\widetilde A$. Let $x \in H$ be an element of the domain of $\widetilde A$ which is not in the domain of $A$. By surjectivity of $A + \mathrm{i}$, there exists $y$ in the domain of $A$ such that $(\widetilde A + \mathrm{i})x = (A + \mathrm{i})y$. Then $x-y$ is nonzero and is annhilated by $\widetilde A+\mathrm{i}$. This is a contradiction because a self-adjoint operator can not have a non-real eigenvalue.