Find the values of the constant $a$ for which the problem $y''(t)+ay(t)=y(t+\pi)$ has a solution with period $2\pi$ which is not identically zero. Also, determine all such solutions
We use a Fourier series solution anstaz, that is $y(t)=\sum_{n\in \mathbb{Z}}c_ne^{int}$. Plugging this into the differential equation, we obtain $$\sum (-n^2+a-(-1)^n)=0 \Rightarrow a=n^2+(-1)^n$$ which is correct (according to the answer). (I'm unsure about the following) Now, since $a$ is a constant, we let $a=m^2+(-1)^m$ to make it clear it does not vary over the sum. Then, by the Fourier series ansatz, $$-n^2+m^2+(-1)^m-(-1)^n=0$$ which gives us that $$n=\sqrt{m^2+(-1)^m-(-1)^n}$$ i.e $n_1=\pm m, n_2=\pm\sqrt{m+2},n_3=\pm\sqrt{m-2}$. This is already slightly alarming as $n\in\mathbb{Z}$, but the terms with the square root need not be integers. I would then set the solutions as $y(t)=A_\pm e^{\pm in_1t}+B_\pm e^{\pm in_2t}+C_\pm e^{\pm in_3t}$ (shorthand for the negative terms aswell). However, the answer states that for $a=n^2+(-1)^n$ with $n\neq 0$, the problem has solutions $y(t)=Ae^{int}+Be^{-int}$ and if $a=1$ then $y(t)=constant$. From this, I suspect that we ought to have $n=\pm m$, but we do not. Can anyone elaborate how they obtain the result and where I go wrong? Thanks