Question

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

Answer 1

no constant.....

Answer 2

Not constant.

Answer 3

i mean it doesnt have a constant

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

@nbouscal find one that satisfies the condition.

Answer 8

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

Answer 9

Or at least, it seems to be.

Answer 10

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

Answer 11

it should have either all even powers or all odd powers

Answer 12

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

Answer 13

x^2-x?

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

lol yea h i got it to :P

Answer 18

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

Answer 19

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.

Answer 20

well that was a good qn :)

Answer 21

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

Answer 22

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

Answer 23

http://www.wolframalpha.com/input/?i=p%28x%5E2%29%3Dp%28x%29p%28x%2B2%29

Answer 24

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.

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

I think a proof would involve polynomial factorization.

Answer 31

only answer is

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

***
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

Answer 38

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

Answer 39

am i right guys?

Answer 40

That all looks right to me, mukushla.

Answer 41

Quite right.