Find all polynomials $P (x)$ $P(x)$ with real coefficients for which $P (x) P (2 x^{2}) = P (2 x^{3} + x)$ $P(x)P(2x^2)=P(2x^3+x)$ ?

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Find all polynomials $P (x)$ $P(x)$ with real coefficients for which $P (x) P (2 x^{2}) = P (2 x^{3} + x)$ $P(x)P(2x^2)=P(2x^3+x)$ ?

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Answer 1 · 2016-11-02T09:14:14

$P_{n} (x) = {(1 + x^{2})}^{n}$ $P_n(x) = (1+x^2)^n$ for $n = 0, 1, 2, 3, \dots$ $n=0,1,2,3,cdots$

Explanation:

$P (2 x^{2}) = \frac{P (2 x^{3} + x)}{P (x)}$ $P(2x^2)=(P(2x^3+x))/(P(x))$ is an even function so

$a)$ $a)$ $P (2 x^{3} + x)$ $P(2x^3+x)$ and $P (x)$ $P(x)$ are odd functions
$b)$ $b)$ $P (2 x^{3} + x)$ $P(2x^3+x)$ and $P (x)$ $P(x)$ are even functions

We have also

$P (0) P (0) = P (0)$ $P(0)P(0)=P(0)$ so

$P (0) = 0$ $P(0)=0$ or $P (0) = 1$ $P(0)=1$

If $P (0) = 0$ $P(0)=0$ then $P (x)$ $P(x)$ must be odd or even. If $P (x)$ $P(x)$ is even

let $a_{2 n} x^{2 n}$ $a_(2n)x^(2n)$ the minimum $x$ $x$ power in it's composition.

Then for that term,

$(P_{b} (x) + a_{2 n} x^{2 n}) (P_{b} (2 x^{2}) + a_{2 n} {(2 x^{2})}^{2 n}) = P_{b} (2 x^{3} + x) + a_{2 n} {(2 x^{3} + x)}^{2 n}$ $(P_b(x)+a_(2n)x^(2n))(P_b(2x^2)+a_(2n)(2x^2)^(2n))=P_b(2x^3+x)+a_(2n)(2x^3+x)^(2n)$

considering the lower order powers to the left and to the rigth

$a_{2 n} x^{2 n} a_{2 n} {(2 x^{2})}^{2 n} = a_{2 n} {(2 x^{3})}^{2 n} + \dots + a_{2 n} x^{2 n}$ $a_(2n)x^(2n)a_(2n)(2x^2)^(2n)=a_(2n)(2x^3)^(2n)+cdots+a_(2n)x^(2n)$ which implies

$a_{2 n} = 0$ $a_(2n)=0$

In the same line of reasoning for the case $P (0) = 0$ $P(0)=0$ and $P (x)$ $P(x)$ odd, we can prove that $P (x)$ $P(x)$ such that $P (0) = 0$ $P(0)=0$ cannot be also odd. So

$P (0) = 1$ $P(0)=1$ and $P (x)$ $P(x)$ is even, then

$P_{n} (x) = 1 + n \sum k = 1 a_{2 n} x^{2 n}$ $P_n(x)=1+sum_(k=1)^na_(2n)x^(2n)$

Now taking $P_{1} (x) = 1 + a_{2} x^{2}$ $P_1(x) = 1+a_2x^2$ we have

$(1 + a_{2} x^{2}) (1 + a_{2} {(2 x^{2})}^{2}) = 1 + a_{2} {(2 x^{3} + x)}^{2}$ $(1+a_2x^2)(1+a_2(2x^2)^2)=1+a_2(2x^3+x)^2$

This equality is verified for $a_{2} = 1$ $a_2=1$ so

$P_{1} (x) = 1 + x^{2}$ $P_1(x) = 1 + x^2$

now considering

$P_{2} (x) = 1 + a_{2} x^{2} + a_{4} x^{4}$ $P_2(x)=1+a_2x^2+a_4x^4$

after

$P_{2} (x) P_{2} (2 x^{2}) = P_{2} (2 x^{3} + x)$ $P_2(x)P_2(2x^2)=P_2(2x^3+x)$

solving for $a_{2}, a_{4}$ $a_2, a_4$ we obtain

$P_{2} (x) = 1 + 2 x^{2} + x^{4} = P_{1} {(x)}^{2}$ $P_2(x) = 1 + 2x^2+x^4 = P_1(x)^2$

Finally we can verify that making

$P_{n} (x) = {(1 + x^{2})}^{n}$ $P_n(x)=(1+x^2)^n$

then it is true

$P_{n} (x) P_{n} (2 x^{2}) = P_{n} (2 x^{3} + x)$ $P_n(x)P_n(2x^2)=P_n(2x^3+x)$

which is equivalent to

${(1 + x^{2})}^{n} {(1 + 4 x^{4})}^{n} = {(1 + x^{2} {(2 x^{2} + 1)}^{2})}^{n}$ $(1+x^2)^n(1+4x^4)^n=(1+x^2(2x^2+1)^2)^n$

and also to

$(1 + x^{2}) (1 + 4 x^{4}) = 1 + x^{2} {(2 x^{2} + 1)}^{2}$ $(1+x^2)(1+4x^4)=1+x^2(2x^2+1)^2$

So the solutions are

$P_{n} (x) = {(1 + x^{2})}^{n}$ $P_n(x) = (1+x^2)^n$ for $n = 0, 1, 2, 3, \dots$ $n=0,1,2,3,cdots$

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Find all polynomials $P (x)$ $P(x)$ with real coefficients for which $P (x) P (2 x^{2}) = P (2 x^{3} + x)$ $P(x)P(2x^2)=P(2x^3+x)$ ?

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Explanation:

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Find all polynomials P(x)P(x)P(x) with real coefficients for which P(x)P(2x^2)=P(2x^3+x)P(x)P(2x2)=P(2x3+x)P(x)P(2x^2)=P(2x^3+x)?

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Explanation:

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Find all polynomials $P (x)$ $P(x)$ with real coefficients for which $P (x) P (2 x^{2}) = P (2 x^{3} + x)$ $P(x)P(2x^2)=P(2x^3+x)$ ?