Find the period of $f(x)$ with $f(x)f(y)=f(x+y)+f(x-y)$.

Question

For any real $x$, $\;\;y$
$f(x)f(y)=f(x+y)+f(x-y)$ with $f(1)=1$
Find the period of $f(x)$.

Answer 1

To summary, I am going to show the following result.

Main Theorem: Let $f$ be the function mentioned in OP. If the peroid of $f$ exists, then the peroid of $f$ only has one of the two forms: $\frac{6}{6n+1}$ and $\frac{6}{6n+5}$ for some $n\in \mathbb{N}$.

@Micah have shown the period $f$ "has" the peroids of $2\cos(\frac{6n\pm 1}{3}\pi x)$, i.e., either $\frac{6}{6n+1}$ or $\frac{6}{6n+5}$. In the following, the "only" is shown.

1. Notations and Lemmas.

Notation 1: Let $D$ be the domain of a function $f$. If $f(x+T)=f(x)$ for all $x\in D$, then $T\in D$ is a peroid of $f$. If $T$ is the least positive one, then $T$ is the peroid of $f$.

Notation 2: Let $\mathbb{Z}$ be the set of all integers. The set of nonnegative integers are denoted by $\mathbb{N}$.

We have two lemmas.

Lemma 1: If the period of $f$ exists, denoted by $P$, then the sete of all peroids is $$T=\{t\in \mathbb{R} \mid f(x+t)=f(x), \forall x \}=\{nP\mid n \in \mathbb{Z}\}.$$ Proof: Please see https://math.stackexchange.com/q/1012902 . Q.E.D.

Lemma 2: Let $f$ be the function in OP. Then the followings hold:

1). $6$ is a peroid of $f$, i.e., $f(x+6)=f(x)$ for all $x \in \mathbb{R}$;

2). $f(0)=2, f(1)=1,f(2)=-1,f(3)=-2,f(4)=-1,f(5)=1$;

3). $f(x+d) \ne f(x)$ for some $x \in \mathbb{R}$, $d=1,2,3$;

4). If the peroid of $f$ exists, then it must has the form $\frac{6}{q}$ where $q\in \mathbb{N}$.

Proof: 1) See the OP; 2) Obviously; 3) As you know $f(x+6)=f(x)$, it is neccesary to show that for any $d\mid 6$, $f(x+d)\ne f(x)$. Recall that $f(x+1)=f(x+2)-f(x)$, $f(x+3)=-f(x)$ and $f(1)=1$. 1. For $d=1$. If $f(x+1)=f(x)$ for all $x$, then we have $$f(1)=f(0+1)=f(2)-f(0)=0.$$ It is a contradiction.

2. For $d=2$. If $f(x+2)=f(x)$ for all $x$, then we have $$f(1)=f(0+1)=f(2)-f(0)=0.$$ It is a contradiction.

3. For $d=3$. If $f(x+3)=f(x)$ for all $x$, then we have $$f(x)=0.$$ It is a contradiction.

4) According to Lemma1 and 1) of Lemma2, if the peroid of $f$ exists, then it must has the form $\frac{6}{q}$ where $q\in \mathbb{N}$.

Q.E.D.

2. The Proof of Main Theorem.

1). If $q=6n$, where $n\in \mathbb{N}$, then the peroid is $\frac{6}{6n}=\frac{1}{n}$. By Lemma 1, $1$ is a peroid. It contradicts to 3) of lemma 2.

2). If $q=6n+1$, where $n\in \mathbb{N}$, then the peroid is $\frac{6}{6n+1}$. For example, $f(x)=2\cos(\frac{6n+1}{3}\pi x)$ and the peroid of $f$ is $\frac{6}{6n+1}$.

3). If $q=6n+2$, where $n\in \mathbb{N}$, then the peroid is $\frac{6}{6n+2}=\frac{3}{3n+1}$. By Lemma 1, $3$ is a peroid. It contradicts to 3) of lemma 2.

4). If $q=6n+3$, where $n\in \mathbb{N}$, then the peroid is $\frac{6}{6n+3}=\frac{2}{2n+1}$. By Lemma 1, $2$ is a peroid. It contradicts to 3) of lemma 2.

5). If $q=6n+4$, where $n\in \mathbb{N}$, then the peroid is $\frac{6}{6n+4}=\frac{3}{3n+2}$. By Lemma 1, $3$ is a peroid. It contradicts to 3) of lemma 2.

6). If $q=6n+5$, where $n\in \mathbb{N}$, then the peroid is $\frac{6}{6n+5}$. For example, $f(x)=2\cos(\frac{6n+5}{3}\pi x)$ and the peroid of $f$ is $\frac{6}{6n+5}$.

Q.E.D.

Answer 2

If $f(x)=2\cos\left(\frac{5\pi}{3}x\right)$, then $f$ satisfies the constraints of the problem. But $f(x+6/5)=f(x)$, so $6$ is not the fundamental period of $f$.

In general we could have $f(x)=2\cos\left(\frac{6n \pm 1}{3}\pi x\right)$ for any $n$, so the fundamental period of $f$ can be made arbitrarily small.