For $r$ a complex root of $P(x)$, thus $P(r)=0$, we get $P(r^2) = 0$, hence $r^2$ is a complex root of $P(x)$. Iterating, this yields $r^{2^k}$ are roots, for $k=0,1,\ldots$. When $P$ is not the identically null polynomial (which is indeed one solution), this implies $r=0$ or $|r|=1$. But also $P((r-2)^2) = P(r-2)P(r) = 0$, so $(r-2)^2$ is a root. Then $r-2=0$, absurd (since $r=2$ is not acceptable), or else $|r-2| = 1$. Then $r=0$ is unacceptable, so $|r|=1$, and the only possibility is $r=1$. Hence $P(x) = k(x-1)^n$. Plugging into the relation yields $k(x-1)^n(x+1)^n = k(x^2-1)^n = P(x^2) = P(x)P(x+2) = k(x-1)^n k(x+1)^n$, whence $k=0$ (the case $P \equiv 0$ already discussed) or $k=1$, so $P(x)=(x-1)^n$. The information about the coefficients being real was superfluous; the proof holds for $P(x) \in \mathbb{C}[x]$.

$ P(x)=1 $ is also a solution.

$P(x) = 1$ is included in my general solution $P(x) = (x-1)^n$, for $n=0$, since that family of (all, together with the null polynomial) solutions is $P_0(x) = 1$; $P_1(x) = x-1$; $P_2(x) = x^2 - 2x + 1$; ................................... $P_n(x) = x^n - \binom {n} {1} x^{n-1} + \cdots + (-1)^{n-1} \binom {n} {n-1} x + (-1)^{n}$.

yes,you're right.sorry.

mavropnevma wrote: For $r$ a complex root of $P(x)$, thus $P(r)=0$, we get $P(r^2) = 0$, hence $r^2$ is a complex root of $P(x)$. Iterating, this yields $r^{2^k}$ are roots, for $k=0,1,\ldots$. When $P$ is not the identically null polynomial (which is indeed one solution), this implies $r=0$ or $|r|=1$. But also $P((r-2)^2) = P(r-2)P(r) = 0$, so $(r-2)^2$ is a root. Then $r-2=0$, absurd (since $r=2$ is not acceptable), or else $|r-2| = 1$. Then $r=0$ is unacceptable, so $|r|=1$, and the only possibility is $r=1$. Hence $P(x) = k(x-1)^n$. Plugging into the relation yields $k(x-1)^n(x+1)^n = k(x^2-1)^n = P(x^2) = P(x)P(x+2) = k(x-1)^n k(x+1)^n$, whence $k=0$ (the case $P \equiv 0$ already discussed) or $k=1$, so $P(x)=(x-1)^n$. The information about the coefficients being real was superfluous; the proof holds for $P(x) \in \mathbb{C}[x]$. $P(r)=0$ then r can be complex but $P(x^2)=P(x)\cdot P(x+2)$ holds when x is real... So, can we say that $P(r^2)=P(r)\cdot P(r+2)$ holds for complex numbers also?

yes,you can because the polynomial $ P(x^{2})-P(x)P(x+2) $ has an infinity of roots so it will be identical $ 0 $ what means that all its coefficients will be $ 0 $.

if the polynomial is not fixed we will prove all its roots are equal 1 , let $p(d.cis(\alpha)) = 0$ then if we put $x=d.cis(\alpha)$ so $p(d^{2^k}.cis(2^k.\alpha)) = 0$ so if $d \ne 1,0$ than $P$ have infinite roots. so let $p(a+bi) = 0, \surd (a^2+b^2) = 1 *$ than if we put $x=a-2+bi$ than $p((a-2+bi)^2) = 0 \Rightarrow (a-2)^2 + b^2 = 1 , * \Rightarrow a = 1, b=0$ and if $d=0$ then $f(4) = 0$ and again $p$ have infinite roots So $p(x) = (x - 1)^\alpha$ and it's correct for any $\alpha$ Note: $cis(\alpha)=cos(\alpha)+i.sin(\alpha)$