Let $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfy $$f(x+y)f(x-y)＝(f(x))^2+(f(y))^2-1,\forall x,y\in\mathbb{R}.$$ $\textbf{My question is: Is $f$ a periodic function}?$
In fact, let $y＝0$, then $(f(0))^2＝1$, and we can assume $f(0)＝1$. Then, if there exists $a\ne0$ such that $f(a)＝1$, then $f(x+a)＝f(x),\ \forall x\in\mathbb{R}$. But how can find this $a$?