Think periodically

As we have seen in Mr Cheong's solution and my commentary, a straightforward approch works pretty well here (although it does not show that 10 is the required minimum). Still, for the sake of variety, let's discuss a more conceptual approach.

For a fixed $f(x)$ and a fixed $x$ , consider the sequence $a_n=f(x+n)$ for $n=0,1,2...$ . This sequence satisfies the linear difference equation $a_n=a_{n-1}-a_{n-2}+a_{n-3}-a_{n-4}$ . The characteristic equation is $c^4-c^3+c^2-c+1=0$ , a factor of $c^5+1$ and therefore of $c^{10}-1$ , so that the roots are the primitive 10th roots of unity, $e^{k\pi{i}/5}$ , for $k=1,3,7,9$ . Thus 10 is a period of $a_n$ . In particular, $a_{10}=a_0$ , meaning that $f(x+10)=f(x)$ , showing that 10 is a period of all such functions.

An example of a solution with fundamental period is 10 is $\cos\left(\frac{\pi{x}}{5}\right)$ , while $\cos\left(\frac{3\pi{x}}{5}\right)$ has the shorter period $\frac{10}{3}$ .

Rearranging the given equation, we have:

$\begin{aligned} f(x+2) & = f(x+1) - f(x)+f(x-1)-f(x-2) \\ f(x+3) & = f(x+2) - f(x+1)+f(x)-f(x-1) = - f(x-2) \\ f(x+4) & = - f(x-2) - f(x+2)+f(x+1)-f(x) = - f(x-1) \\ f(x+5) & = - f(x-1) + f(x-2)+f(x+2)-f(x+1) = - f(x) \\ f(x+6) & = - f(x) + f(x-1)-f(x-2)-f(x+2) = - f(x+1) \\ f(x+7) & = - f(x+1) + f(x)-f(x-1)+f(x-2) = - f(x+2) \\ f(x+8) & = - f(x+2) + f(x+1)-f(x)+f(x-1) = f(x-2) \\ f(x+9) & = f(x-2) + f(x+2)-f(x+1)+f(x) = f(x-1) \\ f(x+10) & = f(x-1) - f(x-2)-f(x+2)+f(x+1) = f(x) \\ f(x+11) & = f(x) - f(x-1)+f(x-2)+f(x+2) = f(x+1) \\ f(x+12) & = f(x+1) - f(x)+f(x-1)-f(x-2) = f(x+2) \end{aligned}$

Therefore, $P = 12-2 = \boxed{10}$

Re-write the equation as

$\displaystyle f(x)=f(x+1)+f(x-1)-f(x+2)-f(x-2)$

Then use this to substitute for $f(x-1)$ :

$f(x+2)+f(x)+f(x-2)=f(x+1)+\left[f(x)+f(x-2)-f(x+1)-f(x-3)\right]$

Everything cancels except for one term on each side

$\displaystyle f(x+2)=-f(x-3)$

which implies $f(x)=-f(x+5)$ , and from there it's easy to see that $f(x+10)=f(x)$ .

This means that f(x) must repeat every 10 units, so there cannot exist a higher $P$ . There cannot exist a lower $P$ since it's simple to construct a function of period 10.

$f\left( x+2 \right) +f\left( x \right) +f\left( x-2 \right) =f\left( x+1 \right) +f\left( x-1 \right) . . . eq 1$

$x\rightarrow x+1$

$f\left( x+3 \right) +f\left( x+1 \right) +f\left( x-1 \right) =f\left( x+2 \right) +f\left( x \right) . . . eq 2$

add $eq 1$ and $eq 2$

$f\left( x+3 \right) +f\left( x-2 \right) =0 . . . eq 3$

$x\rightarrow x+5$

$f\left( x+8 \right) +f\left( x+3 \right) =0 . . . eq 4$

subtract $eq4$ from $eq 3$

$f\left( x-2 \right) =f\left( x+8 \right)$

$x\rightarrow x+2$

$f\left( x \right) =f\left( x+10 \right)$

$P=\huge\boxed { 10 }$