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

Nov 2, 2016

${P}_{n} \left(x\right) = {\left(1 + {x}^{2}\right)}^{n}$ for $n = 0 , 1 , 2 , 3 , \cdots$

#### Explanation:

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

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

We have also

$P \left(0\right) P \left(0\right) = P \left(0\right)$ so

$P \left(0\right) = 0$ or $P \left(0\right) = 1$

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

let ${a}_{2 n} {x}^{2 n}$ the minimum $x$ power in it's composition.

Then for that term,

$\left({P}_{b} \left(x\right) + {a}_{2 n} {x}^{2 n}\right) \left({P}_{b} \left(2 {x}^{2}\right) + {a}_{2 n} {\left(2 {x}^{2}\right)}^{2 n}\right) = {P}_{b} \left(2 {x}^{3} + x\right) + {a}_{2 n} {\left(2 {x}^{3} + x\right)}^{2 n}$

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

${a}_{2 n} {x}^{2 n} {a}_{2 n} {\left(2 {x}^{2}\right)}^{2 n} = {a}_{2 n} {\left(2 {x}^{3}\right)}^{2 n} + \cdots + {a}_{2 n} {x}^{2 n}$ which implies

${a}_{2 n} = 0$

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

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

${P}_{n} \left(x\right) = 1 + {\sum}_{k = 1}^{n} {a}_{2 n} {x}^{2 n}$

Now taking ${P}_{1} \left(x\right) = 1 + {a}_{2} {x}^{2}$ we have

$\left(1 + {a}_{2} {x}^{2}\right) \left(1 + {a}_{2} {\left(2 {x}^{2}\right)}^{2}\right) = 1 + {a}_{2} {\left(2 {x}^{3} + x\right)}^{2}$

This equality is verified for ${a}_{2} = 1$ so

${P}_{1} \left(x\right) = 1 + {x}^{2}$

now considering

${P}_{2} \left(x\right) = 1 + {a}_{2} {x}^{2} + {a}_{4} {x}^{4}$

after

${P}_{2} \left(x\right) {P}_{2} \left(2 {x}^{2}\right) = {P}_{2} \left(2 {x}^{3} + x\right)$

solving for ${a}_{2} , {a}_{4}$ we obtain

${P}_{2} \left(x\right) = 1 + 2 {x}^{2} + {x}^{4} = {P}_{1} {\left(x\right)}^{2}$

Finally we can verify that making

${P}_{n} \left(x\right) = {\left(1 + {x}^{2}\right)}^{n}$

then it is true

${P}_{n} \left(x\right) {P}_{n} \left(2 {x}^{2}\right) = {P}_{n} \left(2 {x}^{3} + x\right)$

which is equivalent to

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

and also to

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

So the solutions are

${P}_{n} \left(x\right) = {\left(1 + {x}^{2}\right)}^{n}$ for $n = 0 , 1 , 2 , 3 , \cdots$