Of course, nothing linear works. Interestingly, every \(f^n(x)\) can be expressed in terms of \(f(x)\) and \(x\). Let's see what this expression is and use \(x \to f(x)\) \[f^3(x) = 2(2f(x) - x - 2) - f(x) - 2 = 3f(x) - 2x - 6\] \[f^n(x) = nf(x) - (n-1)x - n(n-1)\] Okay. \(f\) is injective. If \(f(x) - x = n\), then \(f^{n-1}(x) = x\). \[f^n(x) - xf(x) = (x + n)(f(x) - n + 1)\] Oh bah! \[(f(f(x)) - f(x)) - (f(x) - x) = - 2\] Let \(g(x) = f(x) - x\). \[g(g(x) + x) - g(x) = -2\]