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

[tdenise]

Homework Statement

Find all the polynomials $P(x)$ for which
$$P(x^2+2x+3)=[P(x+3)]^2$$

Homework Equations

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).

Please help me solve this question. Thanks.

 

[the_wolfman]

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

I'd start by solving progressively harder problems.

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

Next let's 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.

 

[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.

 

[tdenise]

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

I'd start by solving progressively harder problems.

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

Next let's 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.

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

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.

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

 

[SammyS]

Homework Statement

Find all the polynomials $P(x)$ for which
$$P(x^2+2x+3)=[P(x+3)]^2$$

Homework Equations

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).

Please help me solve this question. Thanks.

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.

 

[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?

 

[SammyS]

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?

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

 

[haruspex]

Can you please elaborate?

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?