7
$\begingroup$

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)

$\endgroup$
  • $\begingroup$ 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
14
$\begingroup$

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
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.