How to find all polynomials $P(x)$ such that $P(x^2+2x+3)=^2$?

Can someone please show me how to:

Find all polynomials $P(x)$ such that $P(x^2+2x+3)=[P(x+3)]^2$?

I’ve tried substitiuting $x=0,1$. Can’t seem to figure it out. The square on the RHS is confusing me. Thanks.

(P.S. I’m not that familiar with questions of this type)

Solutions Collecting From Web of "How to find all polynomials $P(x)$ such that $P(x^2+2x+3)=^2$?"

Let $Q(x) = P(x+2)$ and $y = x + 1$, we have:

$$Q(y^2) = P(y^2+2) = P(x^2+2x+3) = P(x+3)^2 = Q(y)^2$$

It is then clear $Q(y) = y^n \Leftrightarrow P(x) = (x-2)^n, n \in \mathbb{N}$
is an obvious family of solutions for the functional equation.

By brute force matching of coefficients, it is easy to check for small $n$
(I have checked up to $n = 3$), this is the only solution for $P$ with degree $n$.

Let use prove that this is indeed the case.

Let $Q_1(y), Q_2(y)$ be two polynomial solutions of degree $n > 1$ for the functional equation:
$$Q(y^2) = Q(y)^2\tag{*}$$

By comparing the leading coefficients of $Q_1(y)$ and $Q_2(y)$, we know both of them are monic. This means their difference $U(y) = Q_1(y) – Q_2(y)$ is a polynomial of degree
at most $n-1$. If $U(y)$ is not identically zero, then by comparing degrees on both sides of:

$$U(y^2) = Q_1(y^2) – Q_2(y^2) = U(y)(Q_1(y) + Q_2(y))$$

We get a contradiction that $2 \deg{U} = \deg{U} + n \Leftrightarrow \deg{U} = n$.

From this, we can conclude $Q(y) = y^n \Leftrightarrow P(x) = (x-2)^n$ is the only solution of degree $n$ for corresponding functional equations.

UPDATE Related mathematical topics

Let $R(y)$ be the polynomial $y^2$, the functional equation $(*)$ can be rewritten as:

$$Q\circ R = R\circ Q$$

i.e. the two polynomials $Q$ and $R$ commute under functional composition. There is
a classification theorem which will help us to attack this type of functional equation involving commuting pair of polynomials.

First we need to define the concept of equivalence between 2 pairs of polynomials.
Let $(f_1, g_1)$ and $(f_2, g_2)$ be any two pairs of polynomials, we will call them
equivalent if we can find a linear polynomial $l(x) = ax + b, a \ne 0$ such that:

$$f_1 = l^{-1} \circ f_2 \circ l \quad\text{ and }\quad g_1 = l^{-1} \circ g_2 \circ l$$

In 1922, Ritt proved following theorem:

Let $f$ and $g$ be commuting polynomials. Then the pair $(f,g)$ is
equivalent to one of the following pairs:

  1. $x^m$ and $\epsilon x^n$ where $\epsilon^{m-1} = 1$.
  2. $\pm T_m(x)$ and $\pm T_n(x)$, where $T_m$ and $T_n$ are Chebyshev polynomials.
  3. $\epsilon_1 h^{\circ k}(x)$ and $\epsilon_2 h^{\circ l}(x)$, where $\epsilon_1^q = \epsilon_2^q = 1$ and $h(x)$ is a polynomial of the
    form $x H(x^q)$ and $h^{\circ 1} = h$, $h^{\circ 2} = h\circ h$,
    $h^{\circ 3} = h \circ h \circ h$, and so on.

If you have two commuting polynomials $f$ and $g$, then aside from the trivial case where $f$ and $g$ are functional iterate of a single underlying polynomial $h$. then up to equivalence, $f$ and $g$ can only be simple powers $x^m$ or Chebyshev polynomials $T_m(x)$.

Apply this to our function equation $(*)$. $R$ has the form of a simple power $y^2$,
So $Q(y)$ itself have to be a simple power as we have proved above.

When $R$ is something more complicated, isn’t equivalent to a simple power or Chebyshev polynomial or functional iterate of other polynomial of $h$, then the only solution for
$Q$ are functional iterates $R^{\circ k}$ of $R$.

References

  1. Ritt’s paper Permutable rational functions
  2. V. Prasolov’s book Polynomials