# Looking for P(x)

## Homework Statement

Find all polynomials P(x), such that P(x² - 2x) = [P(x-2)]²

## Homework Equations

Polynomial equations

## 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?

## Homework Statement

Find all polynomials P(x), such that P(x² - 2x) = [P(x-2)]²

## Homework Equations

Polynomial equations

## 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'(y^2+2y)(y+1) = P'(y)P(y)
I am not sure if I derivated correctly, though

Yet, that's it. Any ideas?

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.

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:
haruspex
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Good call, let deg(P(x)) = n, then
2(n-1) + 1 = n(n-1)

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

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

The answer in my cousin's book seems to be

$(x+1)^{n}$

I really thought the answer would be a monster Oo

haruspex
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Why P(x) = Q(x+c)? Did you mean, if the function were periodic of period c, like P(x) = P(x + c)?
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?
By the way, Could you say if I differentiated correctly at post #3? Shouldn't I get an integer n?
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.

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.
Oh, I see now, thanks

haruspex
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Oh, I see now, thanks
Did you find the value of c in my Q()? I think I can help you find how to solve the problem.

Let me retry the problem

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

haruspex
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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
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?

haruspex