Determine all polynominals $\displaystyle P(x)$ (real coeffitients of course) such that for arbitrary number $\displaystyle x$ there is:

$\displaystyle P(x^{2})\times P(x^{3}) = (P(x))^{5}$

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- Sep 1st 2007, 07:54 AMUnunuquantiumPolynominal
Determine all polynominals $\displaystyle P(x)$ (real coeffitients of course) such that for arbitrary number $\displaystyle x$ there is:

$\displaystyle P(x^{2})\times P(x^{3}) = (P(x))^{5}$ - Sep 2nd 2007, 08:42 AMThePerfectHacker
Here is an partial solution. We shall determine all real polynomial $\displaystyle P(x)$ having real zeros that satisfy this.

First $\displaystyle P(x)=C$ implies $\displaystyle C^2 = C \implies C=1\mbox{ or }C=0$.

Now say $\displaystyle P(x)$ is non-constant so $\displaystyle P(x) = (x-a_1)(x-a_2)...(x-a_n)$. Where $\displaystyle a_i$ are not necessrily distinct. (Note $\displaystyle P(x)$ must be a monic polynomial. This is easily to see. Just expand LHS and RHS to the leading term).

Thus,

$\displaystyle (x^2-a_1)...(x^2-a_n)(x^3-a_1)...(x^3-a_n) = (x-a_1)^5...(x-a_n)^5$.

Now here is the result that we need: The equation $\displaystyle x^3 - b = 0$ has complex zeros if and only if $\displaystyle b\not = 0$.

Look at the LHS at the factors $\displaystyle x^3 - a_i$. These has complex zeros if $\displaystyle a_i\not = 0$. But if this is the case then the RHS cannot have the same zeros because it has only real solutions. A contradiction! Thus, $\displaystyle a_i = 0$ for all $\displaystyle 1\leq i \leq n$. And hence the only possibility is $\displaystyle P(x) = (x-0)(x-0)...(x-0) = x^n$.

And this works because if $\displaystyle P(x)=x^n$ we have $\displaystyle (x^2)^n (x^3)^n = (x^n)^5$.

Hence, $\displaystyle P(x) = 0\mbox{ or }1\mbox{ or }x^n$. - Oct 19th 2007, 09:20 AMUnunuquantium
I have some questions to this solution:

1. I cannot see why $\displaystyle P(x)$ is monic polynominal - I mean I see that it must have first coefficient equal to 1, but I cannot see why it must have integer coefficients!

2. I do not understand that $\displaystyle x^{3}-b$ have complex zeros when $\displaystyle b$ is not equal to $\displaystyle 0$.

3. I do not understand why you mean that RHS has got only real zeros! - It has when $\displaystyle P(x)$ is monic??(why??) For example $\displaystyle x^{2}+x+1$ is monic polynominal - but is does not have real solution. That is why you cannot say that $\displaystyle a_{i}$ are ALL real zeros of $\displaystyle P(x)$

4. Why do you say that it is only a partial solution - maybe I did not said the $\displaystyle x$ is not real? So I am sorry, $\displaystyle x$ is REAL.

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Mayby someone has another solution?? - Oct 19th 2007, 09:49 AMThePerfectHacker
The word "monic" means the first coefficient is 1 it does not say anything about the other coefficients.

Note: I made a mistake, I should have been more careful. If the first coefficient is $\displaystyle A$ then this equality leads to $\displaystyle (Ax^2+...)(Ax^3+....)=(A^5x^5+....) \implies (A^2x^5+...)=(A^5x^5+...)$. Thus, $\displaystyle A^2 = A^5 \implies A^2(A^3-1)=0$. We omit the solution $\displaystyle A=0$ because we are working with a polynomial of non-zero coefficient. Thus, $\displaystyle A^3 - 1 = 0$ and hence $\displaystyle A=1,\zeta , \zeta^2$ where $\displaystyle \zeta$ is the 3rd-root of unity.

Hence the first coeffcient can be $\displaystyle 1,\zeta , \zeta^2$. But once you factor that coefficient you have a monic polynomial (the coefficients are not necessarily integers).

Quote:

2. I do not understand that $\displaystyle x^{3}-b$ have complex zeros when $\displaystyle b$ is not equal to $\displaystyle 0$.

Quote:

4. Why do you say that it is only a partial solution?????

Quote:

3. I do not understand why you mean that RHS has got only real zeros! - It has when $\displaystyle P(x)$ is monic - but I cannot see it, as I have written above.

So with my minor mistake fixed the solutions should be this time (for polynomials having zero reals) $\displaystyle 0,1,\zeta,\zeta^2,x^n,\zeta x^n,\zeta^2 x^n$.

EDIT: Ignore what I just said on the bottom. I forgot that you said "polynomials with real coefficients". So my original solution is correct. - Oct 19th 2007, 10:04 AMUnunuquantium
#1 - ok.

#2 :), my stupid mistake, ignore it, i did not see sth, sorry

#3What if we have $\displaystyle x^{2}+x+1$ - this polynominal has only complex solutions, and: it's coefficients are real, and x is real, and it is monic.

??

#4 ok

I just do not understand why you have written:

$\displaystyle P(x) = (x-a_1)(x-a_2)...(x-a_n)$

maybe some $\displaystyle a_{i}$ are complex?? - Oct 19th 2007, 10:49 AMThePerfectHacker
- Oct 19th 2007, 11:36 AMUnunuquantium
Ok, sorry, I forgot what was the question exactly.

Thank you very much - Oct 19th 2007, 12:26 PMCaptainBlack
Let $\displaystyle P(x)=a_nx^n + a_mx^m + .. ,\ n>m \ge 0$ be the highest order

terms, then:

$\displaystyle

(a_nx^n + a_mx^m + ..)^5=(a_nx^{2n} + a_mx^{2m} + ..)(a_nx^{3n} + a_mx^{3m} + .. )

$

expanding these as far as the two highest order terms on each side:

$\displaystyle

a_n^5x^{5n} + 5 a_n^4 a_m x^{4n+m}+..= a_n^2x^{5n}+a_na_mx^{3n+2m}+..

$

So by equating exponents of the second heighest order terms we conclude

that $\displaystyle n=m$ a contradiction. So $\displaystyle P(x)=a x^n$ if $\displaystyle n \ge 1$, and then we must

have $\displaystyle a^5=a^2$, so if $\displaystyle a$ is real $\displaystyle a=1$.

This all assumed that $\displaystyle n>1$, so we need also to check

if $\displaystyle P(x)\equiv 1$ and $\displaystyle P(x) \equiv 0$ are solutions, which they are.

RonL - Oct 19th 2007, 12:29 PMCaptainBlack