ALG-B3-M07-P018 Even Multiplicativity

First analyse nonzero roots, then the parity of the degree.

The constant solutions are \(0\), \(1\). Let \(P\) be nonconstant. If \(r\ne0\) is a root, then \(P(r^2)=P(r)P(-r)=0\), again producing an infinite chain of roots. Thus the only root is \(0\), so \(P(x)=ax^n\). Substitution gives \(a x^{2n}=a^2(-1)^n x^{2n}\). Hence \(a=a^2(-1)^n\). Since \(a e0\), \(a=(-1)^n\). The answer is \(P(x)=x^{2k}\) and \(P(x)=-x^{2k+1}\), together with \(0\), \(1\).

Strong task on roots and the sign of the leading coefficient.