In the previous post, we have to show that the solution of
\begin{equation}
f(x+1)+f(x-1)=\sqrt{2}f(x)
\end{equation}
is periodic and we are asked to find the period.
First evaluate the equation in \(x\to x+1\) and \(x\to x-1\),
\begin{equation}
f(x+2)+f(x)=\sqrt{2}f(x+1)
\end{equation}
\begin{equation}
$$f(x)+f(x-2)=\sqrt{2}f(x-1)\;.
\end{equation}
Now we add this two equations
\begin{equation}
f(x+2)+f(x-2)+2f(x)=\sqrt{2}[f(x+1)+f(x-1)]
\end{equation}
and replace Eq. (1) in the right side of this last equation
\begin{equation}
f(x+2)+f(x-2)+2f(x)=2f(x)
\end{equation}
which is equivalent to
\begin{equation}
f(x+2)=-f(x-2)
\end{equation}
which proves that \(f(x+8)=f(x)\), i.e. \(f\) is periodic with period 8.
