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Mathematics 44 Online
OpenStudy (anonymous):

What is the general form of a polynomial p(x) such that that p(x^2)=p(x)p(x+2)

OpenStudy (anonymous):

no constant.....

OpenStudy (anonymous):

Not constant.

OpenStudy (anonymous):

i mean it doesnt have a constant

OpenStudy (anonymous):

x^2+x is a polynomial without a constant, but does not satisfy this problem.

OpenStudy (anonymous):

Even simpler, x is a polynomial without a constant, but does not satisfy this problem.

OpenStudy (anonymous):

It could have a constant. I need the general form of the solution.

OpenStudy (anonymous):

@nbouscal find one that satisfies the condition.

OpenStudy (anonymous):

I am attempting to, but this is a rather tricky problem.

OpenStudy (anonymous):

Or at least, it seems to be.

OpenStudy (anonymous):

The zero polynomial is obviously a trivial solution. I have not yet found a non-trivial solution.

OpenStudy (anonymous):

it should have either all even powers or all odd powers

OpenStudy (anonymous):

@eliassaab, are you sure there are non-trivial solutions?

OpenStudy (anonymous):

x^2-x?

OpenStudy (anonymous):

naah i missed it....shd be near that is wrong all even and odd thing may not be true sry

OpenStudy (anonymous):

let x=0 then p(0)=p(0)p(2) then p(0)=0 or p(2)=1 if P(0)=0 then let x=-2 it gives p(4)=p(-2)p(0)=0 P(16)=p(4)Pp(6)=0 .... so p(0)=/=0 and p(2)=1

OpenStudy (anonymous):

Okay, I'm dumb. I was doing the problem wrong. p(x)=x-1 is a non-trivial solution. p(x^2) there is x^2-1 and p(x)p(x+2) is (x-1)(x+1)=x^2-1

OpenStudy (anonymous):

lol yea h i got it to :P

OpenStudy (anonymous):

the trick is aw we go on for higher powers things will not be cancelled coz of lame constants and powers should be consecutive so

OpenStudy (anonymous):

It's aggravating to me that p(x)=x-1 is the first polynomial that I tried, and because I computed it wrong I threw it out.

OpenStudy (anonymous):

well that was a good qn :)

OpenStudy (anonymous):

Still is a good question. I still do not know what the general form of the solution is.

OpenStudy (anonymous):

Yup. \(p(x)=x-1\) is the only solution, I am pretty sure.

OpenStudy (anonymous):

p(x)=x-1 if the polynomial is of the form x+a with a constant. That much I am certain of. It being the only solution I have not yet been convinced of.

OpenStudy (anonymous):

I do wish that wolf would show steps on that. It's hard to trust wolf because it is so often wrong =/

OpenStudy (anonymous):

Avinash's reasoning convinced me. It's vague.

OpenStudy (anonymous):

edit: It's vague, but his point makes sense.

OpenStudy (anonymous):

Even if p(x)=x-1 is the only solution, I would like to be able to prove that formally.

OpenStudy (anonymous):

I can formally prove that it is the only solution of the form x+a, but past that, I do not know.

OpenStudy (anonymous):

I think a proof would involve polynomial factorization.

OpenStudy (anonymous):

only answer is \[\huge P(x)=(x-1)^n\]

OpenStudy (anonymous):

Confirmed that \((x-1)^2\) is a solution. Nice work, mukushla.

OpenStudy (anonymous):

\[ \begin{align} p(x^2)&=(x^2-1)^n\\ p(x)p(x+2)&=(x-1)^n(x+1)^n\\ &=((x-1)(x+1))^n\\ &=(x^2-1)^n \end{align} \]

OpenStudy (anonymous):

so, mukushla's answer is correct for all \(n\).

OpenStudy (anonymous):

I don't see a proof that this is the only solution, however.

OpenStudy (anonymous):

i proved that p(0)=/=0 now Suppose p(a) = 0 for some complex number a. Then p(a^2) = p(a^4) =· · · = 0. then |a| = 1. Similarly, if a is any such root, then P((a −2)^2) = P(a − 2) P(a) = 0. So (a − 2)^2 is also a root. It follows that |a − 2| = 1. there is only 1 complex numbers such that |a| = 1 and |a −2 | = 1 and thats a=1 so the only root of P(x) is x=1 then we have \[P(x)=a(x-1)^n\] put this in the equation that gives a^2=a ---> a=1 because leading coefficient of 2 sides must be equal so u have \[\large P(x)=(x-1)^n\] for any positive integer n

OpenStudy (anonymous):

*** Suppose p(a) = 0 for some complex number a. Then p(a^2) = p(a)p(a+2) =0 p(a^4) = p(a^2)p(a^2+2) =0 ............. . However, a polynomial can only have a finite number of roots. So a must be a root of unity. Therefore, |a| = 1

OpenStudy (anonymous):

Similarly, we can prove that |a + 2| = 1

OpenStudy (anonymous):

am i right guys?

OpenStudy (anonymous):

That all looks right to me, mukushla.

OpenStudy (anonymous):

Quite right.

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