Find polynomials $P(x)P(x-3) = P(x^2)$

Question

Find all polynomials $P \in \Bbb R[x]$ such that $P(x)P(x-3) = P(x^2) \quad \forall x \in \Bbb R$

I have a solution but I'm not sure about that. Please check it for me.

It is easy to see that $P(x) = 0, P(x) = 1$ satisfied.

Consider $P(x) \neq c$ :

We have $P(x+3)P(x) = P((x+3)^2)$

If $a$ is a root of $P(x) $ then $a^2$ and $(a+3)^2$ are roots of $P(x)$.

If $|a| > 1$, then $ |a^2| = |a||a| > |a| \Rightarrow P(x)$ has infinitely many roots.

If $0 < |a| < 1$, then $|a^2| = |a||a| < |a| \Rightarrow P(x)$ has infinitely many roots.

If $a = 0 \Rightarrow (a+3)^2 = 9$ is a root of $P(x)$, so $P(x)$ has infinitely many roots.

Hence, every root $a$ of $P(x)$ has $|a| = 1$. So $1 = |(a+3)^2| = (|a+3|^2) \Rightarrow |a+3|=1$.

We have $a + 3 = \cos\alpha + i\sin\alpha + 3 \Rightarrow (\cos\alpha+3)^2 + \sin^2\alpha = 1 \Rightarrow \cos\alpha = -\frac{3}{2}\ (\text{impossible}).$

$P(x) = 0, P(x) = 1$ are all the results.

Answer 1

Considering

$$ P(x)P(x-b)=P(x^2),\ \ \ b \ne 0\ \ \ \ (1) $$

As long as $P(x^2)$ is even then $P(x)P(x-b)$ is also even then

$$ P(b)P(0) = P(b^2) $$

because making $x = b$ in $(1)$ we have $P(b)P(0) = P(b^2)$, As this is for a generic $b$ we have $P(x)P(0) = P(x^2)$

or

$$ P(x)P(0) = P(x^2) = P(-x)P(-(x+b)) $$

which implies that $P(x)$ is even as well as $P(x+b)$

This should be true for all $b \ne 0$ so the solution is $P(x) = C_0$

Finally, according to $C_0^2=C_0$ we conclude that $C_0 = 0$ or $C_0 = 1$

Answer 2

Your approach is sound and the arguments besides the last case $|a|=1$ are clear. Here is a slightly more stringent variant, which does not bring anything new, besides maybe the modified argument for the case $|a|=1$.

We are looking for all polynomials $P\in\mathbb{R}[x]$ such that \begin{align*} P(x)P(x-3)=P(x^2)\qquad \forall x\in\mathbb{R}\tag{1} \end{align*}

Proof:

• Constant Polynomial $P$: At first we look at polynomials $P(x)=c$ which are constant. From (1) we obtain \begin{align*} c^2&=c\\ c(c-1)&=0\tag{2} \end{align*} Since the equation (2) is solved iff $c=0$ or $c=1$, these are the only solutions when $P$ is a constant polynomial.

In the following we consider all possible cases when $P$ has a root $a\in\mathbb{R}$. We divide the cases into

\begin{align*} |a|>1,\qquad 0<|a|<1,\qquad a=0,\qquad a=-1\quad \text{and} \quad a=1 \end{align*} which cover all possibilities.

• Root $P(a)=0$ with $|a|>1$:

  Let $a\in\mathbb{R}$ with $|a|>1$. We get from (1) \begin{align*} P(a^2)=P(a)P(a-3)=0\cdot P(a-3)=0\tag{3} \end{align*}

  From (3) we conclude that there is an infinite sequence $(a^{2n})_{n\geq 0}$ of pairwise different roots of $P$. Thus, $P$ has infinitely many roots and it follows $P$ is the zero-polynomial: $P=0$.

In the same way we can treat the next case.

• Root $P(a)=0$ with $0<|a|<1$:

  Let $a\in\mathbb{R}$ with $0<|a|<1$. We get from (1) \begin{align*} P(a^2)=P(a)P(a-3)=0\cdot P(a-3)=0\tag{4} \end{align*}

  From (4) we conclude as we did in the case before that there is an infinite sequence $(a^{2n})_{n\geq 0}$ of pairwise different roots of $P$. Thus, $P$ has infinitely many roots and it follows $P$ is the zero-polynomial: $P=0$.

Three more cases to follow:

• Root $a=0$:

  Let $a=0$ with $P(0)=0$. We get from (1) setting $x=3$

\begin{align*} P(9)=P(3)P(0)=P(3)\cdot 0=0 \end{align*}

We see that $9$ is also a root of $P$. We are now in the case with root $|a|>1$ and conclude $P=0$.

• Root $a=-1$:

  Let $a=-1$ with $P(-1)=0$. We get from (1) setting $x=2$

\begin{align*} P(4)=P(2)P(-1)=P(2)\cdot 0=0 \end{align*}

We see that $4$ is also a root of $P$. We are now in the case with root $|a|>1$ and conclude $P=0$.

• Root $a=1$:

  Let $a=1$ with $P(1)=0$. We get from (1) setting $x=4$

\begin{align*} P(16)=P(4)P(1)=P(4)\cdot 0=0 \end{align*}

We see that $16$ is also a root of $P$. We are now in the case with root $|a|>1$ and conclude $P=0$.

Conclusion: We have covered all types of different polynomials which fulfill the relationship (1) and conclude that $\color{blue}{P=0}$ and $\color{blue}{P=1}$ are the only solutions of (1).$\qquad\qquad\qquad\Box$

Note: When making rigorous proofs we should avoid phrases like: It is clear that ..., It is easy to see that .... It is much more preferable to prove even seemingly simple things, just to be sure that we don't overlook anything.

Answer 3

You point out that if $a$ is a root then $a^{2}$ is a root and $a+3$ is a root.

You don't need to divide into three cases to determine that means either infinitely or zero roots.

By induction $a + 3k; k \in \mathbb N$ are roots. If $j\ne k$ then $a+3j \ne a + 3k$ so if there is one root $a$ there are at least countably infinite distinct roots. So there are either zero roots, of infinitely many roots.

Magnitude is irrelevant[1]

Only trivial constant polynomials have infinite or zero (complex) roots so that means $P(x) = c$ for some $c$ where $c^2 = P(x)P(x-3) =P(x^2) = c$.

[1]

Although if $a$ is a root then $a +3$ is a root. And if $|a+3| < 1$ then $|a+3| + |a|=|a+3| + |-a| \ge |(a+3)+(-a)| =|3|=3$ so $|a|\ge 3-|a+3| > 2$. So at least one root is $\ge 1$.