# Difference between revisions of "1997 AHSME Problems/Problem 27"

## Problem

Consider those functions $f$ that satisfy $f(x+4)+f(x-4) = f(x)$ for all real $x$. Any such function is periodic, and there is a least common positive period $p$ for all of them. Find $p$. $\textbf{(A)}\ 8\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 32$

## Solution

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 $\boxed{D}$. 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 $f(1) \neq f(1 + 8)$, 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 $f(1) \neq f(1 + 12)$, 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 $f(1) \neq f(1 + 16)$, 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 $f(1)$.