# Looking for P(x)

1. Aug 13, 2014

### Dinheiro

1. The problem statement, all variables and given/known data
Find all polynomials P(x), such that P(x² - 2x) = [P(x-2)]²

2. Relevant equations
Polynomial equations

3. The attempt at a solution
Let x = y+2, then
P(y²+2y) = P²(y)

I tried to analyse by derivating
P'(y²+2y)(2y+2) = 2P(y)P'(y)
P'(y²+2y)(y+1) = P'(y)P(y)
I am not sure if I derivated correctly, though

Yet, that's it. Any ideas?

2. Aug 14, 2014

### gopher_p

I would consider what $\deg\big( P(x)\big)$ says about $\deg\big(P'(y²+2y)(y+1)\big)$ and $\deg\big( P'(y)P(y)\big)$. I'm not 100% certain, but I'm guessing you will get a good constraint on $\deg\big( P(x)\big)$ which will make it easier to just brute force the coefficients.

Also, and this is just a pet peeve of mine, the appropriate verb and its tenses are differentiate, differentiating, and differentiated.

3. Aug 14, 2014

### Dinheiro

Good call, let deg(P(x)) = n, then
2(n-1) + 1 = n(n-1)
n² - 3n +1 = 0??
Did I differentiated correctly?
By the way, not peeve at all, my english is quite rusty and I miss sometimes some english technical terms, thanks for the correction xD

Last edited: Aug 14, 2014
4. Aug 15, 2014

### haruspex

Think again about the right hand side of that equation. (I think the corrected form will be tautology.)

Try some other lines:
- find a value of x for which P can be evaluated (or, at least, for which you can limit the possible values to a small set).
- suppose P has a root, α. What other roots are implied?
- what can you say about a local minimum of P?

Last edited: Aug 15, 2014
5. Aug 15, 2014

### haruspex

... or consider P[x] = Q[x+c]. See if you can find a value of c which yields a simpler equation.

6. Aug 16, 2014

### Dinheiro

Why P(x) = Q(x+c)? Did you mean, if the function were periodic of period c, like P(x) = P(x + c)?

By the way, Could you say if I differentiated correctly at post #3? Shouldn't I get an integer n?
Thanks

7. Aug 16, 2014

### Dinheiro

The answer in my cousin's book seems to be

$(x+1)^{n}$

I really thought the answer would be a monster Oo

8. Aug 16, 2014

### haruspex

No, not periodic:
P(y²+2y) = P²(y)
Q(y²+2y+c) = Q²(y+c)

Can you spot a value of c that makes that interesting?
You differentiated ok, but wrongly evaluated the highest power that would arise on the right hand side. If you do it correctly, you'll find that the equation doesn't tell you anything.

9. Aug 16, 2014

### Dinheiro

Oh, I see now, thanks

10. Aug 17, 2014

### haruspex

Did you find the value of c in my Q()? I think I can help you find how to solve the problem.

11. Aug 17, 2014

### Dinheiro

Let me retry the problem

12. Aug 17, 2014

### Dinheiro

c = 1, so that Q(y²+2y+1) = Q[(y+1)²]
and, therefore,
Q[(y+1)²]=Q²[(y+1)]
Great, haruspex!
Now, Let y+1 = z
Q(z²)=Q²(z)
Now, I got stucked, how can I really prove Q(z) = (z+1)^n from it? I can guess the answer, but I couldn't really demonstrate it

13. Aug 18, 2014

### haruspex

Good progress. Now you're in a better position to try one of my first suggestions. (It might have worked with P(), but it's easier with Q().) Suppose α is a root (not necessarily real). What other roots are implied?

14. Aug 18, 2014

### haruspex

Edit: There's an easier way: just consider the lowest two degree terms in Q, $a_r x^r + a_s x^s$, say. Equate the lowest two terms of Q(x2) with the lowest two of Q2(x).

15. Aug 22, 2014

### Dinheiro

Great! Thanks, haruspex!