For the difference equation \(x_{n+1}=\beta x_n- g(x_n)\) with \(\beta> 1\) and \(g(x)= \text{sign\,}x\) for \(x\neq 0\), \(g(0)= 1\), it is shown: For any \(m\in \mathbb{N}_0\) it has a \(2^m\)-periodic solution. If \(\beta^{2^m(2k+1)}- 2\beta^{2^m(2k-1)}\geq 1\) for \(k\in \mathbb{N}\) and some \(m\in\mathbb{N}\) it has a \((2k+1)2^m\)-periodic solution. In the case \(m= 0\) the condition is also necessary.