If $f(x + 1) + f(x − 1) = f(x), \forall x \in \mathbb{R}$,then how to find $k$ such that $f(x + k) = f(x)$?

Let $f(x)$ be a function such that $f(x + 1) + f(x − 1) = f(x), \forall x \in \mathbb{R}$. Then for what value of $k$ is the relation $f(x + k) = f(x)$ necessarily true for every real $x$?

The answer/solution suggested in my module is like this: "this is a bit involved but can be proved that $k=6$". Could anybody explain me this?

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Just add two consecutive relations: $$f(x+2)+f(x)=f(x+1)$$ $$f(x+3)+f(x+1)=f(x+2)$$ Then you'll get $$f(x+3)+f(x)=0$$ vor every real $x$. You have also $f(x+6)+f(x+3)=0$ for every real $x$, and therefore $f(x)=f(x+6)$ for every real $x$.

Another way. Consider the recurrence relation $x_{n+1}+x_{n-1}=x_n$, $x_0, x_1 \in \Bbb{R}$. Then the general term of the sequence is $$x_n=a \varepsilon^n +b \varepsilon^n$$ where $a,b$ are real numbers determined by $x_0,x_1$ and $\varepsilon$ verifies the characteristic equation $x^2-x+1=0$. This implies that $\varepsilon^3=-1$ and $\varepsilon^6=1$. I think this approach may prove that we can construct some function $f$ for which $k=6$ is the least $k$ for which $f(x)=f(x+6)$ for every $x \in \Bbb{R}$.

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There are many ways to get the desired result. Beni’s first method is about as simple as they come, but a slightly different one occurred to me first:

\begin{align*} 2f(x)&=f(x-1)+f(x)+f(x+1)\\ &=f(x-2)+f(x-1)+2f(x)+f(x+1)+f(x+2)\\ &=f(x-2)+3f(x)+f(x+2)\;, \end{align*}

so $$f(x-2)+f(x)+f(x+2)=0\;.\tag{1}$$ Replacing $x$ by $x+2$, we have $$f(x)+f(x+2)+f(x+4)=0\;.\tag{2}$$ Now just subtract $(1)$ from $(2)$ to get $$f(x+4)-f(x-2)=0\;,$$ or $f(x+4)=f(x-2)$ for all $x\in\Bbb R$, which is of course equivalent to $f(x+6)=f(x)$ for all $x\in\Bbb R$.

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