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)

  • $\begingroup$ Is the RHS $[P(x+3)]^2$ or $P[(x+3)^2]$? $\endgroup$ – ABC Apr 11 '13 at 2:39
  • $\begingroup$ The former. Edited $\endgroup$ – Denise T Apr 11 '13 at 2:39

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 $R$.


  1. Ritt's paper Permutable rational functions
  2. V. Prasolov's book Polynomials
  • $\begingroup$ This seems really complicated for the level we're at. Is there an easier solution? I lost you after "If U(y) is not identically zero, then by comparing degrees on both sides of:" Could you please elaborate on that part? Thanks. $\endgroup$ – Denise T Apr 11 '13 at 18:09
  • $\begingroup$ @DeniseT How about this argument: If $\alpha$ is a root of $Q(y)$, then $Q(y^2) = Q(y)^2$ implies $\pm \sqrt{\alpha}$ are also roots of $Q(y)$. If $Q(y)$ contains any non-zero root, it will lead to a contradiction that $Q(y)$ contains infinite many roots. So $Q(y)$ only has 0 as its root. Since it has to be monic, $Q(y)$ is a simple power $y^n$ for some $n$. $\endgroup$ – achille hui Apr 11 '13 at 18:18
  • $\begingroup$ @DeniseT about comparing the degree. If $U(y)$ is a polynomial of degree $m$, then $U(y^2)$ is a polynomial of degree $2m$. Similarly, if you have two non-zero $V(y), W(y)$, the degree of their product is the sum of the degree: $\deg U(y)W(y) = \deg U(y) + \deg V(y)$. $\endgroup$ – achille hui Apr 11 '13 at 18:22
  • $\begingroup$ Thanks for for helping me out here but I feel really stupid. Your explanation with the root seems easier. I understand the second half but I don't see how $Q(y^2)=Q(y)^2$ implies that $\pm \sqrt{a}$ are also roots. The way I see it: If $a$ is a root of $Q(y)$ then $Q(y)^2=(0)^2=0$. And since $Q(y^2)=Q(y)^2$, $Q(y^2)=0$, meaning that $y^2$ is also a root. So wouldn't the implication be: if $a$ is a root of $Q(y)$ then $a^2$ is also a root. $\endgroup$ – Denise T Apr 11 '13 at 19:09
  • 1
    $\begingroup$ 1) if $a$ is a root, then $Q(\pm\sqrt{a})^2 = Q((\sqrt{a})^2) = Q(a) = 0$, so $\pm\sqrt{a}$ are also roots. 2) The two operations $a \to a^2$ and $a \to \pm \sqrt{a}$ are not the same thing. the first one doesn't guarantee an increase in the number of roots while the second one does. 3) By fundamental of algebra, if $Q(y)$ is a polynomial of degree $n$, then $y$ has exactly $n$ roots (counting multiplicity) over $\mathbb{C}$. If you can show $Q(y)$ doesn't have a root, then it is a non-zero constant. $\endgroup$ – achille hui Apr 12 '13 at 2:44

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