I am trying to find the eigenstates/eigenvalues of the following Hamiltonian $$ \hat{H} = \hbar \omega \Big(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\Big)+A\big(\hat{a}^{\dagger}\hat{a}^{\dagger}+\hat{a}\hat{a}\big)\equiv \hat{H}_1+\hat{H}_2, $$ where $\hat{a}$ and $\hat{a}^{\dagger}$ are the standard ladder operators, with $[\hat{a}, \hat{a}^{\dagger}] = 1$.
My attempt
(1) I tried to check whether the two terms were commuting, to see if the problem admitted a separable solution, but $\big[\hat{H}_1, \hat{H}_2\big]= \hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a}$.
(2) I tried the trial solution (even states) $$ |\phi\rangle = \sum_n c_{n2}|2n\rangle $$ which when substituted gives $$ \hat{H}|\phi\rangle = \sum_n \left[\frac{\hbar \omega}{2}c_{2n}+\hbar \omega (2n) c_{2n}+c_{2n-2}+c_{2n+2}\right]|2n\rangle $$ which gives the infinite system $$ c_{2n}= \left(\frac{\hbar \omega}{2}+2n\hbar \omega\right)c_{2n} +c_{2n-2}+c_{2n+2}. $$ I am not sure if this leads anywhere.
Question
Is there a standard way to approach this problem? Any help?