# How to find find all P(x) for this functional equation

1. Apr 10, 2013

### tdenise

1. The problem statement, all variables and given/known data
Find all the polynomials $P(x)$ for which
$$P(x^2+2x+3)=[P(x+3)]^2$$

2. Relevant equations

3. The attempt at a solution
I don't really know how to solve functional equations systematically. I tried to to find a linear $P(x)$ and found $P(x)=x-2$ through trial & error. I also tried substituting x = 0 and x = (x-3) but that didn't go anywhere.

$$x=0: P(3) = [P(3)]^2$$
$$\Rightarrow{P(3) = {0,1}}$$

$$x=(x-3): P((x-2)^2+2) = [P(x)]^2$$

Note that on the LHS, there is $(x-2)$ (the P(x) I found) but I don't know if that's a coincidence or not. Also the LHS & RHS are ≥ 0 and $P(x)$ must be an "odd degree" polynomial (i.e. the leading term must be an odd power).

2. Apr 11, 2013

### the_wolfman

At first glance this looks like a challenging problem...

I'd start by solving progressively harder problems.

First lets assume that P(y) is a constant. P(y)=c. What values of c satisfy the equation c=c^2 ?

Next lets look at the case where P(y) is linear. P(y) = a y + b. Now we want to find a and b that satisfy the equation a(x^2+2x+3) + b = (a(x+3)+b)^2.

After that go to the quadratic case. Hopefully a few patterns will begin to emerge.

3. Apr 11, 2013

### haruspex

Suppose n ≥ 1 is the max n for which (x-α)n is a factor of P, some root α. Let P(x) = (x-α)nQ(x). What equation can you deduce for Q? Can you deduce from this further roots as functions of α? If so, that potentially leads to an infinite sequence of roots, which would not be allowed, so the task becomes to find those α for which the sequence terminates.

4. Apr 11, 2013

### tdenise

The expansion on the RHS is too long for thee quadratic case and higher to see any pattern.

I don't see how you can deduce anything for Q from that. Can you please elaborate?

5. Apr 11, 2013

### SammyS

Staff Emeritus
You can determine something about the polynomial, P(x), evaluated at a few values of x.

It may be helpful to use a different variable to express $\displaystyle \ P(u^2+2u+3)=(P(u+3))^2\ .$

If $\displaystyle \ u^2+2u+3=u+3\,,\$ then u = 0 or -1 .

This gives that $\displaystyle \ P(3)=(P(3))^2\,, \$ which you already have.

Also, it gives $\displaystyle \ P(2)=(P(2))^2\ .$

What if u = -2 ?

This is far from solved, but gives something to work with.

6. Apr 11, 2013

### tdenise

Thanks SammyS. For u = -2: $P(3) = (P(1))^2$, so
$$(P(3))^2 = (P(1))^2$$

So P(x) is does not have a one to one y to x mapping (I don't remember the name for that). How does that help solve the problem though?

I also noticed that the minimum of $x^2+2x+3$ is 2. Is this relevant?

7. Apr 11, 2013

### SammyS

Staff Emeritus
Well, you do know something about P(3), right? ... It's either 0 or 1 .

8. Apr 11, 2013

### haruspex

I'll start by switching x to x-3 everywhere. This produces an equivalent equation but with a more convenient RHS: P(x2-4x+6) = P(x)2
Let P(x) = (x-α)nQ(x) with n≥1, so (x2-4x+6-α)nQ(x2-4x+6) = P(x2-4x+6) = P(x)2 = (x-α)2nQ(x)2.
so (x2-4x+6-α) divides (x-α)2nQ(x)2.
Under what circumstances does x-α divide (x2-4x+6-α)?
If it does not divide it, what roots can you deduce for Q?