Design a function \(f: \mathbb{Z} \rightarrow \mathbb{Z}\) such that \(f(f(n)) = -n\).
An example is:
\[f(n) = \operatorname{sgn}(n) - n(-1)^n\]Proof
Suppose that \(f(x) = y\) for some \(x\) and \(y\). Then:
\[\begin{align} f(y) = & f(f(x)) = -x \\ f(-x) = & f(f(y)) = -y \\ f(-y) = & f(f(-x)) = x \\ \end{align}\]Therefore continued application of \(f\) results in a cycle: \(x \rightarrow y \rightarrow -x \rightarrow -y \rightarrow x\).
This implies that \(f(0) = 0\) because we require that \(x = f(0) = f(-0) = -x\).
For all other integers we need to partition them into groups of 4, each which contain some \(x\), \(y\) and their negations. To construct such a mapping, let us make the simplest choice of letting \(f(1) = 2\). This then forces the following cycle: \(1 \rightarrow 2 \rightarrow -1 \rightarrow -2 \rightarrow 1\).
Likewise we can let \(f(3) = 4\) and in general for \(k > 0\) we can let \(f(2k-1) = f(2k)\) which forces: \(2k-1 \rightarrow 2k \rightarrow -2k-1 \rightarrow -2k \rightarrow 2k-1\). Thus we can define how the function acts on a number based on whether it is positive or negative and whether it is even or odd:
\[f(n) = \begin{cases} n + 1 & \text{ if } n > 0 \wedge n \text{ is odd} \\ -n + 1 & \text{ if } n > 0 \wedge n \text{ is even} \\ n - 1 & \text{ if } n < 0 \wedge n \text{ is odd} \\ -n - 1 & \text{ if } n < 0 \wedge n \text{ is even} \end{cases}\]When \(n < 0\) we are always adding 1, and when \(n > 0\) we are always subtracting \(1\). Thus the above simplifies to:
\[f(n) = \begin{cases} \operatorname{sgn}(n) + n & \text{ if } n \text{ is odd} \\ \operatorname{sgn}(n) - n & \text{ if } n \text{ is even} \end{cases}\]Now \((-1)^n = 1\) when \(n\) is even, and \((-1)^n = -1\) when \(n\) is odd. This allows us to simplify the above to our solution: \(f(n) = \operatorname{sgn}(n) - n(-1)^n\)
By construction it satisfies \(f(f(n)) = -n\) for \(n \ne 0\), and we can evaluate \(f(0)\) to check that it indeed holds for \(0\) as well. Thus our solution is correct.