What are the eigenvector's of the $\hat a^2$ operator? Since $\hat a^2$ and $\hat a$ commute, then one of the eigenvectors of $\hat a^2$ will be, the coherent state $|\alpha\rangle$. Are there others states as well?
 A: Such states do exist. The simplest examples are the symmetric cat states
\begin{align}
|\psi_+\rangle & = \frac{|\alpha\rangle + {|-\alpha} \rangle }{\sqrt{2}} \\
|\psi_-\rangle & = \frac{|\alpha\rangle - |{-\alpha} \rangle }{\sqrt{2}}, 
\end{align}
tough any similar linear combination will work. 
A: The eigenstates of $\hat a^2$ are called Barut-Girardello coherent states.  They are naturally associated with the $\mathfrak{su}(1,1)$ algebra since
$$
\hat K_+=\frac{1}{2}(\hat a^\dagger)^2\, ,\qquad \hat K_-=\frac{1}{2}\hat a^2\, ,\qquad \hat K_0=\frac{1}{4}(\hat a\hat a^\dagger+\hat a^\dagger\hat a)
$$
close on the $\mathfrak{su}(1,1)\sim \mathfrak{so}(2,1)$ algebra.  
This algebra is closely connected to many radial-type differential equations, and so applications include various types of radial potentials.  An example is

Popov, Dusan. "Barut-Girardello coherent states of the pseudoharmonic oscillator." Journal of Physics A: Mathematical and General 34.25 (2001): 5283.

(Note they are not the usual Perelomov-type coherent states for $\mathfrak{su}(1,1)$.  Whereas $[\hat a^\dagger,\hat a]$ is basically the unit operator, $[\hat K_+,\hat K_-]$ is diagonal but NOT proportional to the unit, so the BG coherent states do not have the group properties that the Perelomov coherent states have.)
A: The question is - Is there an eigenstate of ${\hat a}^2$ that is not an eigenstate of ${\hat a}$?
To determine this, suppose there exists a state $|\psi\rangle$ which is an eigenstate of ${\hat a}^2$ with eigenvalue $\lambda$ but is not an eigenstate of ${\hat a}$. Then,
$$
|\phi\rangle \equiv {\hat a} |\psi\rangle \not\propto | \psi\rangle 
$$
Note that $|\phi \rangle$ is also an eigenvector of ${\hat a}^2$ with the same eigenvalue $\lambda$. We also have ${\hat a} |\phi\rangle = \lambda |\psi\rangle$. Now, define two new states (for $\lambda \neq 0$)
$$
|\psi_\pm\rangle = |\phi\rangle \pm \sqrt{\lambda} |\psi\rangle
$$
We can immediately find 
$$
a |\psi_\pm\rangle = \pm \sqrt{\lambda}   |\psi_\pm\rangle
$$
Thus, we find that if there exists a state which is an eigenvector of ${\hat a}^2$ but not of ${\hat a}$, there must exist another state satisfying the same property. One can then always find two linear combinations of the states which are eigenvectors of both ${\hat a}^2$ and ${\hat a}$. 
