Problem
ALG-B3-M08-P016 Third Polynomial Iteration
#16
★★★★★ Level 5 of 5
Find all \(P\in\mathbb R[x]\) such that \(P^3(x)=x\), where \(P^3=P\circ P\circ P\).
Compare degrees, then solve the linear case.
If \(P\) is constant, impossible. Let \(\deg P=d\). Then \(\deg P^3=d^3\), while the right-hand side has degree \(1\), so \(d=1\). Let \(P(x)=ax+b\). The coefficient of \(x\) in the third iterate is \(a^3\), so \(a^3=1\), hence over \(\mathbb R\), \(a=1\). Then \(P(x)=x+b\), and \(P^3(x)=x+3b\). Thus \(b=0\). The answer is \(P(x)=x\).
Polynomial cycle of length \(3\) over \(\mathbb R\).