How many Possible Functions?

Any function $f(x)$ such that $f(x)^2=1$ for all $x$ takes the values $1,-1$ only. If such a function is to be discontinuous at the integers only, it must be constant on each interval $(n,n+1)$ where $n$ is an integer.

Let $X(n)$ be the collection of $\{1,-1\}$ -valued functions on $[0,n]$ that are discontinuous at precisely the integers $\{0,1,2,...,n\}$ , and let $Y(n)$ be the collection of $\{1,-1\}$ -valued functions on the open interval $(0,n)$ that are discontinuous at precisely the integers $\{1,2,...,n-1\}$ . We note immediately that $|X(n)| = |Y(n)|$ , since any element of $Y(n)$ must extend uniquely to an element of $X(n)$ (the value at $0$ must be opposite to the value on $(0,1)$ , and the value at $n$ must be opposite to the value on $(n-1,n)$ ).

For any element $f$ of $Y(n)$ , let us consider the extensions of $f$ that belong to $Y(n+1)$ . If $g \in Y(n+1)$ is such that $g(x) = f(x)$ for all $0<x<n$ , then

• If the value of $g$ on $(n,n+1)$ is the same as the value of $f$ on $(n-1,n)$ , then the value of $g$ at $n$ must take the opposite value,
• if the value of $g$ on $(n,n+1)$ is the opposite to the value of $f$ on $(n-1,n)$ , then $g$ can take either value $1$ or $-1$ at $n$ .

Thus there are $3$ elements of $Y(n+1)$ that are extensions of $f$ . Of course every element of $Y(n+1)$ is an extension of an element of $Y(n)$ , and hence we deduce that $|Y(n+1)| = 3|Y(n)|$ . Since $|Y(1)| = 2$ , we deduce that $|X(n)| = |Y(n)| = 2 \times 3^{n-1} \hspace{2cm} n \ge 1$ Thus we deduce that $|X(5)| =\boxed{162}$ .