# Polynominal

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• September 1st 2007, 07:54 AM
Ununuquantium
Polynominal
Determine all polynominals $P(x)$ (real coeffitients of course) such that for arbitrary number $x$ there is:
$P(x^{2})\times P(x^{3}) = (P(x))^{5}$
• September 2nd 2007, 08:42 AM
ThePerfectHacker
Quote:

Originally Posted by Ununuquantium
Determine all polynominals $P(x)$ (real coeffitients of course) such that for arbitrary number $x$ there is:
$P(x^{2})\times P(x^{3}) = (P(x))^{5}$

Here is an partial solution. We shall determine all real polynomial $P(x)$ having real zeros that satisfy this.
First $P(x)=C$ implies $C^2 = C \implies C=1\mbox{ or }C=0$.

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

Thus,
$(x^2-a_1)...(x^2-a_n)(x^3-a_1)...(x^3-a_n) = (x-a_1)^5...(x-a_n)^5$.

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

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

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

Hence, $P(x) = 0\mbox{ or }1\mbox{ or }x^n$.
• October 19th 2007, 09:20 AM
Ununuquantium
I have some questions to this solution:

1. I cannot see why $P(x)$ 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 $x^{3}-b$ have complex zeros when $b$ is not equal to $0$.

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

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

--------
Mayby someone has another solution??
• October 19th 2007, 09:49 AM
ThePerfectHacker
Quote:

Originally Posted by Ununuquantium
1. I cannot see why $P(x)$ 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!

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 $A$ then this equality leads to $(Ax^2+...)(Ax^3+....)=(A^5x^5+....) \implies (A^2x^5+...)=(A^5x^5+...)$. Thus, $A^2 = A^5 \implies A^2(A^3-1)=0$. We omit the solution $A=0$ because we are working with a polynomial of non-zero coefficient. Thus, $A^3 - 1 = 0$ and hence $A=1,\zeta , \zeta^2$ where $\zeta$ is the 3rd-root of unity.
Hence the first coeffcient can be $1,\zeta , \zeta^2$. But once you factor that coefficient you have a monic polynomial (the coefficients are not necessarily integers).

Quote:

2. I do not understand that $x^{3}-b$ have complex zeros when $b$ is not equal to $0$.
Do you something about complex numbers? The solution to the equation $x^3 = b$ is $\sqrt[3]{b} , \sqrt[3]{b}\zeta,\sqrt[3]{b}\zeta$ where $\zeta$ is the 3rd-root of unity. Now if $b \not = 0$ these solutions are non-zero and hence complex (provided that $b\in \mathbb{R}$).

Quote:

4. Why do you say that it is only a partial solution?????
We are considering all the polynomials that have only real zeros. The polynomials that have complex zeros are not being considered here.

Quote:

3. I do not understand why you mean that RHS has got only real zeros! - It has when $P(x)$ is monic - but I cannot see it, as I have written above.
By what is answered to #4 the polynomial $P(x)$ here has only real zeros. So $P^5(x)$ (the RHS) has only real zeros.

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

EDIT: Ignore what I just said on the bottom. I forgot that you said "polynomials with real coefficients". So my original solution is correct.
• October 19th 2007, 10:04 AM
Ununuquantium
#1 - ok.
#2 :), my stupid mistake, ignore it, i did not see sth, sorry
#3What if we have $x^{2}+x+1$ - 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:
$P(x) = (x-a_1)(x-a_2)...(x-a_n)$
maybe some
$a_{i}$ are complex??
• October 19th 2007, 10:49 AM
ThePerfectHacker
Quote:

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

I just do not understand why you have written:
$P(x) = (x-a_1)(x-a_2)...(x-a_n)$
maybe some
$a_{i}$ are complex??

[/QUOTE]
What I did does not apply to this polynomial. Because I was only doing ones with real zeros not complex.
• October 19th 2007, 11:36 AM
Ununuquantium
Ok, sorry, I forgot what was the question exactly.

Thank you very much
• October 19th 2007, 12:26 PM
CaptainBlack
Quote:

Originally Posted by Ununuquantium
Determine all polynominals $P(x)$ (real coeffitients of course) such that for arbitrary number $x$ there is:
$P(x^{2})\times P(x^{3}) = (P(x))^{5}$

Let $P(x)=a_nx^n + a_mx^m + .. ,\ n>m \ge 0$ be the highest order
terms, then:

$<br /> (a_nx^n + a_mx^m + ..)^5=(a_nx^{2n} + a_mx^{2m} + ..)(a_nx^{3n} + a_mx^{3m} + .. ) <br />$

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

$<br /> a_n^5x^{5n} + 5 a_n^4 a_m x^{4n+m}+..= a_n^2x^{5n}+a_na_mx^{3n+2m}+..<br />$

So by equating exponents of the second heighest order terms we conclude
that $n=m$ a contradiction. So $P(x)=a x^n$ if $n \ge 1$, and then we must
have $a^5=a^2$, so if $a$ is real $a=1$.

This all assumed that $n>1$, so we need also to check
if $P(x)\equiv 1$ and $P(x) \equiv 0$ are solutions, which they are.

RonL
• October 19th 2007, 12:29 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Hence, $P(x) = 0\mbox{ or }1\mbox{ or }x^n$.

This is overdone as $P(x)=1$ is a case of $P(x)=x^n$.

RonL
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