1. ## A functional equation

Find all polynomials $p(x) \in \mathbb{C}[x]$ which satisfy the identity $p(2x-x^2)=(p(x))^2.$

2. $2x-x^2=1-(1-x)^2$, so if $p(x)=(1-x)^n$, we see that
$p(2x-x^2)=[1-(2x-x^2)]^n=(1-2x+x^2)^n=(1-x)^{2n}=(p(x))^2$.

--Kevin C.

3. Originally Posted by TwistedOne151
$2x-x^2=1-(1-x)^2$, so if $p(x)=(1-x)^n$, we see that
$p(2x-x^2)=[1-(2x-x^2)]^n=(1-2x+x^2)^n=(1-x)^{2n}=(p(x))^2$.

--Kevin C.
$p(x)=0$ is also a solution. the question is that whether or not these are the only possible solutions?

4. OK, let's offer this to the snake and see if it strikes.

Using the functional equation, first show that if $a$ is a root of $p(z)=0$ then so are $2a-a^2$ and $1\pm\surd(1-a)$.

Also, since $p(z)=0$ is a polynomial equation it has only a finite number of roots.

Now let $m=\max_{p(z)=0}|1-z|$. By the previous assertion this is well-defined. Aim to show that $m=0$ and therefore that $1$ is the only possible root of $p(z)=0$.

1. Assume $m>1$. If $z$ is a root with $|1-z|=m$, let $w=2z-z^2$. Then $w$ is a root with $|1-w|=|1-z|^2=m^2>m$, contradicting the choice of $m$.

2. Assume $0. Choose a root $z$ with $|1-z|=m$ and let $w=1\pm\surd(1-z)$. Then $w$ is a root with $|1-w|=\surd|1-z|=\surd m>m$, contradiction.

3. Assume $m=1$. Choose a root $z$ with $|1-z|=1$. If $z=0$ is a root then $z=2$ is also a root. Therefore assume WLOG that $z=1-\mathrm e^{\mathrm i\theta}$ for some $\theta$ in $(0,2\pi)$.

Then $w=1-\surd(1-z)=1-\mathrm e^{\mathrm i\theta/2}$ is a root. Similarly $1-\mathrm e^{\mathrm i\theta/4}$ is a root, and $1-\mathrm e^{\mathrm i\theta/8}$, etc.

Continue in this way to show the existence of an infinite number of distinct roots of the polynomial equation $p(z)=0$, another contradiction.

Therefore $m=0$ is the only possibility, and consequently $z=1$ is the only possible root. Therefore write $p(z)=k(1-z)^m$ where $m\in\mathbb N\cup\{0\}$ and $k$ is a constant.

Use the functional equation to get $k=k^2$, and so $k=0$ or $k=1$.