Question

What is the general form of a polynomial p(x) of degree 2m such that p(x^2)=p(x)p(x+1)

Answer 1

\[P(x)=(x^2-x+1)^m\]

Answer 2

Are you sure?

Answer 3

wait a sec

Answer 4

this time i am thinking that \(p(0)\) unless this is a constant function

Answer 5

@mukushla
\[
\left\{p\left(x^2\right)=x^4-x^2+1\\p(x)
p(x+1)=x^4+x^2+1\right\}
\]
in your solution for m=1

Answer 6

since as before since \(p(0)=p(0)p(1)\implies p(0)=0\) or \(p(1)=1\) but if
\(p(1)=1\) then since \(p(1)=p(1)p(2)\) we have \(p(1)=0\) which it is not or \(p(2)=p(1)\)

Answer 7

The only constant solutions are p(x) = 0 and p(x) = 1 for all x. So
we restrict consideration to the case where p(x) is nonconstant. now for x = 0 and x = −1 the equation becomes
p(0) = p(0) p(1), p(1) = p(−1) p(0) .
then 0 is a root if and only if 1 is a root. so i think P(x) is in th form
\[P(x)=a x^l (1-x)^k Q(x)\]

Answer 8

but we have to find Q(x) !!!

Answer 9

zero is a root yes? which makes 1 a root as well

Answer 10

let x=-1
p(1) = p(−1) p(0)

Answer 11

right, and since \(p(0)=0\) 1 is a root also

Answer 12

unless my logic was faulty above

Answer 13

yes thats right

Answer 14

now i think we can put the P(x) in the equation to get the Q(x)

Answer 15

let me do that

Answer 16

coefficient of largest exponent of x must be equal on 2 sides so a=a^2 then a=1

Answer 17

\[ax^{2l} (x − 1)^l (x + 1)^l Q(x^2) = a^2 x^{l+k} (x + 1)^l (x − 1)^k Q(x) Q(x + 1)\] i got rid of the large so i could read it

Answer 18

any way to use that the sum of the coefficients is zero?

Answer 19

also as your argument before, if \(r\) is a root, then so is \(r^2\) limiting the possible zeros

Answer 20

i dont know how to use that (sum of the ...=0)

Answer 21

it seems l=k but i cant prove it

Answer 22

looks similar to previous one
only possible roots are 0, 1, \(-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i\) by what you wrote in previous post
that is, if \(r\) is a root, so is \(r^2, r^4, ...\)

Answer 23

the problem is 0,1
in the previous one 0,1 were not roots

Answer 24

i guess i meant only other possible roots

Answer 25

maybe these are the only two
which is what i suspect
a quick calculation tells me for example that \(p(x)=x(1-x)\) works whereas
\[p(x)=x(1-x)(x^2+x+1)=x-x^4\] does not

Answer 26

oh yes sorry
u mean to find all possible roots for this one

Answer 27

i am off to the movies
love to know what the solution is, so i'll check back later

Answer 28

ok

Answer 29

ok i want to prove 0,1 are only roots of such polynomials
similar to previous one
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 − 1)^2) = P(a − 1) P(a) = 0.
So (a − 1)^2 is also a root. It follows that |a − 1| = 1.
there are only 2 complex numbers such that |a| = 1 and |a − 1| = 1 .
\[a=\frac{1}{2}\pm \frac{\sqrt{3}}{2} i\]
but unlike previous one
\[a^2=-\frac{1}{2}\pm \frac{\sqrt{3}}{2} i \neq a\]
that means only ossible roots are 0 and 1 so
\[P(x)=c x^l (x-1)^k\]
put this in the equation we get
\[cx^{2l}(x−1)^k(x+1)^k=c^2x^{l+k}(x+1)^l(x−1)^k \ \ \ \ \ \ (I)\]
leading coefficient of 2 sides must be equal then
c=c^2 ----> c=1

from (I)
\[(x(x+1)(x-1))^{l-k}=1 \rightarrow \rightarrow l=k=m\]
so the only answer is

\[\large P(x)=x^m (x-1)^m\]

Answer 30

@eliassaab
@satellite73

Answer 31

@mukushla Excellent solution.

Answer 32

tnx