Find all real polynomial solutions of a functional equation

Question

Let f be a real polynomial

a)

$f(x)f(x+1)=f(x^2+x+1)$

b)

$f(x)f(2x^2)=f(2x^3+x)$

Answer 1

a) $$f(x)f(x+1)=f(x^2+x+1)$$

Note that $f(x)=0$ works. If $f(x) \not =0$, note that $f(x)=x^2+1$ works. Thus if we write $f(x)=(x^2+1)^ng(x)$, where $n \geq 0, n \in \mathbb{Z}$, and $x^2+1$ is not a factor of the polynomial $g(x)$, then $g(x)$ also satisfies the equation $g(x)g(x+1)=g(x^2+x+1)$. Note that $g(x) \not =0$.

If $g(x)=c \not =0$ is a constant, then $c^2=c$, so $c=1$. In fact, $g(x)=1$ works.

Otherwise $g(x)$ has a root $\alpha$.

Note that if $\alpha$ is a root of $g(x)$, then $g(\alpha)=0$, so $g(\alpha^2+\alpha+1)=g(\alpha)g(\alpha+1)=0$ and $g(\alpha^2-\alpha+1)=g(\alpha-1)g(\alpha)=0$, so $\alpha^2+\alpha+1$ and $\alpha^2-\alpha+1$ are also roots of $g(x)$. If $\pm i$ is a root of $g(x)$, then $(\pm i)^2+ \pm i+1=\pm i$ and $(\pm i)^2-(\pm i)+1= \mp i$ are roots of $g(x)$, so $x^2+1$ is a factor of $g(x)$, a contradiction. Therefore $i, -i$ are not roots of $g(x)$.

We shall show that at least one of $|\alpha^2+\alpha+1|, |\alpha^2-\alpha+1|$ is $>|\alpha|$. Assume on the contrary that $|\alpha^2+\alpha+1| \leq |\alpha|$ and $|\alpha^2-\alpha+1| \leq |\alpha|$. Then by triangle inequality $2|\alpha| \geq |\alpha^2+\alpha+1|+|\alpha^2-\alpha+1|=|\alpha^2+\alpha+1|+|-\alpha^2+\alpha-1| \geq |2\alpha|=2|\alpha|$. Thus equality holds, so $|\alpha^2+\alpha+1|=|-\alpha^2+\alpha-1|=|\alpha|$ and $\alpha^2+\alpha+1, -\alpha^2+\alpha-1, \alpha$ have the same argument, so $\alpha^2+\alpha+1=-\alpha^2+\alpha-1=\alpha$. Thus $\alpha^2+1=0$, so $\alpha= \pm i$, a contradiction.

Therefore at least one of $|\alpha^2-\alpha+1|, |\alpha^2+\alpha+1|$ is $>|\alpha|$. We now construct an infinite sequence of roots of $g(x)$ with increasing modulus as follows: $\alpha_1=\alpha$, and $\alpha_{n+1}=\alpha_n^2 \pm \alpha_n+1$, where the sign is chosen to make $|\alpha_n^2 \pm \alpha_n+1|>|\alpha|$. (If both signs work, randomly choose one of them) This implies that $g(x)$ has infinitely many roots, a contradiction.

Therefore $g(x)=1$, so $f(x)=(x^2+1)^n$. To conclude, $f(x)=0$ and $f(x)=(x^2+1)^n$ are the only polynomial solutions for the given equation.

b) $$f(x)f(2x^2)=f(2x^3+x)$$

Note that $f(x)=0$ works. If $f(x) \not =0$, note that $f(x)=x^2+1$ works. Thus if we write $f(x)=(x^2+1)^ng(x)$, where $n \geq 0, n \in \mathbb{Z}$, and $x^2+1$ is not a factor of the polynomial $g(x)$, then $g(x)$ also satisfies the equation $g(x)g(2x^2)=g(2x^3+x)$. Note that $g(x) \not =0$.

If $g(x)=c \not =0$ is a constant, then $c^2=c$, so $c=1$. In fact, $g(x)=1$ works.

Otherwise $g(x)$ has a root $\alpha$.

Note that substituting $x=0$ gives $g(0)^2=g(0)$, so $g(0)=0, 1$.

If $g(0)=0$, then write $g(x)=x^ah(x), a \in \mathbb{Z}^+, h(0) \not =0$, then $x^ah(x)(2x^2)^ah(2x^2)=(2x^3+x)^ah(2x^3+x)$. Thus $2^ax^{2a}h(x)h(2x^2)=(2x^2+1)^ah(2x^3+x)$, so substituting $x=0$ gives $h(0)=0$, a contradiction.

Thus $g(0)=1$, so $0$ is not a root of $g(x)$.

If $\alpha$ is a root of $g(x)$, then $g(\alpha)=0$, so $g(2\alpha^3+\alpha)=0$, so $2\alpha^3+\alpha$ is also a root of $g(x)$. If $\pm i$ is a root of $g(x)$, then $2(\pm i)^3 \pm i=\mp i$ is also a root of $g(x)$, so $x^2+1$ is a factor of $g(x)$, a contradiction. Thus $\pm i$ are not roots of $g(x)$.

If $\alpha$ is a root of $g(x)$ for some $|\alpha| \geq 1$, then note that by triangle inequality $|2\alpha^2+1|+1=|2\alpha^2+1|+|-1| \geq |2\alpha^2| \geq 2$, so $|2\alpha^3+\alpha|=|\alpha||2\alpha^2+1| \geq |\alpha| \geq 1$. If $|2\alpha^3+\alpha|=|\alpha|$, then we must have $|2\alpha^2+1|=1$, so equality must hold in the triangle inequality above, so $2\alpha^2+1$ must be a negative real number, so $\alpha=\pm i$, a contradiction. Therefore $|2\alpha^3+\alpha| \not =|\alpha|$, so $|2\alpha^3+\alpha|>|\alpha| \geq 1$.

We can thus construct an infinite sequence of roots of $g(x)$ with increasing modulus as follows: $\alpha_1=\alpha, \alpha_{n+1}=2\alpha_n^3+\alpha_n$, so $|\alpha_{n+1}|>|\alpha_n|>1$. Therefore $g(x)$ has infinitely many roots, a contradiction.

Thus all roots $\alpha$ of $g(x)$ satisfy $0<|\alpha|<1$. Finally note that $1=|f(0)|=\prod_{g(\alpha)=0}{|\alpha|^{m(\alpha)}}<1$ (where $m(\alpha)$ is the multiplicity of the root $\alpha$), a contradiction.

Therefore the only solution is $g(x)=1$, so $f(x)=(x^2+1)^n$. To conclude, $f(x)=0$ and $f(x)=(x^2+1)^n$ are the only polynomial solutions for the given equation.