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We have:

$\begin{aligned} f(x) + f(x+2) &= f(x+1) \\ f(x+1) + f(x+3) &= f(x+2) \end{aligned}$

Adding them together, we have:

$\begin{aligned} f(x) + f(x+1) + f(x+2) + f(x+3) &= f(x+1) + f(x+2) \\ f(x) + f(x+3) &= 0 \end{aligned}$

By substitution $x \gets x+3$ , we have:

$f(x+3) + f(x+6) = 0$

The two equations together imply:

$\begin{aligned} f(x) + f(x+3) &= f(x+3) + f(x+6) \\ f(x) &= f(x+6) \end{aligned}$

This implies that $6$ is a period. To show that this is the fundamental period, we have to show one function satisfying the equation where its period is $6$ . The first function that comes to mind is a rather messy one:

Let $g : \mathbb{Z} \to \mathbb{R}$ be defined as:

$g(n) = \begin{cases} 0 &\text{if } n \equiv 0, 3 \pmod{6} \\ 1 &\text{if } n \equiv 1, 2 \pmod{6} \\ -1 &\text{if } n \equiv 4, 5 \pmod{6} \end{cases}$

Now define $f$ as $f(x) = g \left( \lfloor x \rfloor \right)$ for all real $x$ . It can be verified that $f$ satisfies the equation and the period of this function is equal to the period of $g$ , which is $6$ , proving that $\boxed{6}$ is its fundamental period.

Fancy method: For a fixed $x$ , we have a Fibonacci style recursive equation $f(x+2)=f(x+1)-f(x)$ whose general real solution is $f(x+n)= a\cos(n\pi/3)+b\sin(n\pi/3),$ so that $f(x+6)=f(x)$

Pedestrian method: If $f(x)=a$ and $f(x+1)=b$ then $f(x+2)=b-a, f(x+3)=-a, f(x+4)=-b, f(x+5)=a-b$ and $f(x+6)=a=f(x)$

A simple example is $f(x)=\sin\left(\frac{\pi{x}}{3}\right)$ , with fundamental period $\boxed{6}$ .

$f(x-1) + f(x+1) = f(x).$

Plugging $x+1$ into $f(x)$ , we see $f(x+1) = f(x) + f(x+2).$

Thus, $f(x-1) + f(x) + f(x+2) = f(x)$

$\; \Rightarrow f(x-1) + f(x+2) = 0 \\ \Rightarrow f(x) + f(x+3) = 0 \\ \Rightarrow f(x) = - f(x+3) \\ \Rightarrow f(x+3) = -f(x+6) \\ \Rightarrow f(x) = -(-f(x+6)) \\ \Rightarrow f(x) = f(x+6).$

We see that 6 could be the fundamental period. We must now verify that 1, 2, and 3 are not guaranteed periods of $f$ .

Consider $f(x) =cos(\pi x/3).$

$cos(0) \not= cos(2\pi/3)$ , so 2 is not necessarily a period.

$cos(0) \not= cos(\pi)$ , so 3 is not necessarily a period either.

Since 2 is not necessarily a period, 1 does not need to be one either, as then since 2 is a multiple of 1, 2 would have to also be a period.

Thus 6 is the fundamental period of $f$ .

just take f(x)=A a^x .and solve for A and a

Define the operator $E$ , such that $Ef(x)=f(x+1)$ . (Recall that an operator is something that turns a function into another function. $E$ shifts $f$ over by one.) Also, have $E^0=I$ be the identity operator, defined by $Ig=g$ for any function $g$ .

We are given $E^{-1}f+Ef=f$ . Rearranging, we get $Ef-f+E^{-1}f=0$ . In other words, $(E-I+E^{-1})f=0$ , or — multiplying everything by $E$ on the left: $(E^2-E+I)f=0$ We want to find the period of $f$ . Now, if $f$ had a period of $n$ , we would have $f(x+n)=f(x)$ , or $f(x+n)-f(x)=0$ , or: $(E^n-I)f=0$ So our problem boils down to: What polynomial $E^n-I$ has $E^2-E+I$ as a factor?

Now, let's pretend — for just a second — that $E$ is an ordinary variable, rather than an operator. So we want to find a polynomial $E^n-1$ that's a multiple of $E^2-E+1$ . To do so, notice that the roots of $E^2-E+1$ are $\frac12\pm i\frac{\sqrt3}2$ . These are sixth roots of unity! That means that $E^2-E+1$ is a factor of $E^6-1$ ! So, $\boxed6$ is our answer.

Summary: Your equation is $E^{-1}f+Ef=f$ , and the solutions to $E^{-1}+E=1$ are sixth roots of unity. Thus, the answer is $\boxed6$ .

If you're not sure how that worked, notice that we can perform long division to show that $(E^4+E^3-E-1)(E^2-E+1)=E^6-1$ . Thus, starting from our equation from earlier: $(E^2-E+I)f=0$ and multiplying everything by $E^4+E^3-E-I$ on the left, we get: $(E^4+E^3-E-I)(E^2-E+I)f=0$ $(E^6-I)f=0$ and thus $f(x+6)-f(x)=0$ , and $6$ is our period.