smc

Recall that $p$ is the fundamental period of function $f$ iff $p$ is the smallest positive $p$ such that $f(x)=f(x+p)$ for all $x$. In this case, we know that $f(x+4)+f(x-4)=f(x)$. Plugging in $x+4$ in for $x$ to get the next equation in the recursion, we also get $f(x+8)+f(x)=f(x+4)$. Adding those two equations gives $f(x+8)+f(x-4)=0$ after cancelling out common terms. Again plugging in $x+4$ in for $x$ in that last equation (in order to get $f(x)$), we find that $f(x)=-f(x+12)$. Now, plugging in $x+12$ for $x$, we get $f(x+12)=-f(x+24)$. This proves that $f(x)=f(x+24)$, so there is a period of $24$, which gives answer $\boxed{24}$. We now eliminate answers $A$ through $C$. Let $f(x)=x$ for $x\in \{1,2,3,4,5,6,7,8\}$. Plugging in $x=5$ into the initial equation gives $f(9)+f(1)=f(5)$, which implies that $f(9)=4$. Since $f(1)\ne f(1+8)$, the function does not have period $8$. Continuing, $f(10)=f(6)-f(2)=4$, $f(11)=f(7)-f(3)=4$, and $f(12)=f(8)-f(4)=4$. Once we hit $f(13)$, we have $f(13)=f(9)-f(5)=4-5=-1$. Since $f(1)\ne f(1+12)$, the function does not have period $12$. Finally, $f(14)=f(10)-f(6)=4-6=-2$, $f(15)=f(11)-f(7)=4-7=-3$, $f(16)=f(12)-f(8)=4-8=-4$, and $f(17)=f(13)-f(9)=-1-4=-5$. Since $f(1)\ne f(1+16)$, the function does not have period $16$. To confirm that our original period works, we may see that $f(18)=-6$, $f(19)=-7$, $f(20)=-8$, $f(21)=-4$, $f(22)=-4$, $f(23)=-4$, and $f(24)=-4$. Finally, $f(25)=1$, which is indeed the same as $f(1)$.