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The full question is to solve $f(x)f(2x^2) = f(2x^3+x)$, prove there are at most one solution per degree of $f(x)$ and then find all the solutions. (Here $f(x)$ is assumed to be polynomial.)

So far I have proven that $(x^2+1)^n$ works for even degree polynomials, and I'm pretty near certain there are no odd solutions.

Is the following a valid argument for there only being at most 1 solution per degree? Let $K$ be the degree of $f(x)$ then $f(x)f(2x^2) = f(2x^3+x)$ produces $3k+1$ linearly independent equations with only $k+1$ independent variables. Therefore for any degree $K$ polynomial there can be at most 1 solution.

Finally, is there any elegant way to prove that no odd solutions exist?

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    $\begingroup$ Is $f:\mathbb{R}\to\mathbb{R}$? $\endgroup$ – Zero Jan 24 '15 at 19:30
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    $\begingroup$ Hint: If $x$ is root of $f$, so is $2x^3 + x$. $\endgroup$ – gtrrebel Jan 24 '15 at 19:42
  • $\begingroup$ @Zero yes, it is $f:\mathbb{R}\to\mathbb{R}$ $\endgroup$ – Sinister Jan 24 '15 at 21:13
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    $\begingroup$ @ Lukas - it's a polynomial, it has to be continuous $\endgroup$ – Asier Calbet Jan 24 '15 at 22:55
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    $\begingroup$ Well, if $P(x_0)=0$ we'd also have $P(2x^3_0+x_0)=0$. If $x_0$ is real and $\neq0$, that would give an infinity of zeros, that's impossible for a polynomial of any degree. But polynomials of odd degree always have a real zero. So we'd have to exclude that, but $P(0)=0$ would imply that $P(2x^3+x)$ is divisible by $x\cdot x^2$, so $P(x)$ must be divisible by $x^3$, and so on. $\endgroup$ – Professor Vector Jan 11 '18 at 13:04
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As observed in other answers, the leading coefficient of $f$, and the constant term both have to equal $1$ (if $f$ is not identically $0$). Now assume that $f$ has complex root with absolute value greater than 1. Let $z$ one of them with greatest absolute value. Now $2z^3 + z$ is also a root, by triangle inequality $$ |2z^3 + z| = |z||2z^2 + 1| \geq |z|(|2z^2| - 1) > |z|(2 - 1) = |z|, $$ contradicting the assumption. So all the roots have absolute value $\leq 1$. But the product of the absolute values of roots is $1$ (the constant term is $1$), so all of them are of absolute value one.

Again using the previous inequality with $|z| = 1$ instead of $|z| > 1$, we get that for all the roots $$ |2z^2 + 1| = |2z^2| - 1, $$ which is quite easily only possible if $z^2 = -1$ so $z = \pm i$. So only possible roots are $\pm i$ and if $i$ is, so is $2i^3 + i = -i$ and vice versa.

Lastly, to ensure that multiplicities of $i$ and $-i$ are equal: if $f$ has roots, it has factor $z^2 + 1$ by our observation, and factoring that out we see that the resulting polynomial also satisfies the functional equation. So indeed, only solutions are constant polynomials $0$ and $1$ and $(x^2+1)^k$ for positive integer $k$.

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  • $\begingroup$ Great answer! Also shows how useful complex numbers can be. $\endgroup$ – Akiva Weinberger Jan 24 '15 at 23:22
  • $\begingroup$ Perfect! I was thinking in this direction, wanting to generalise my ideas on the root with maximum value in my answer below by thinking in terms of absolute value like you did - I came to your conclusion that the absolute value of the roots would have to be less than or equal to 1, but was stuck there. Your insight that since their product is 1 their absolute values must all be 1 is perfect and exactly what I was missing. Well done! $\endgroup$ – Asier Calbet Jan 25 '15 at 20:13
  • $\begingroup$ Yep, I was stuck there for a while too, I mean with the possibility of "smaller" roots. It's kinda nice that you can solve that problem just by looking those two coefficients, which is not that hard. $\endgroup$ – gtrrebel Jan 25 '15 at 21:11
  • $\begingroup$ No, it should be correct as it is. $\endgroup$ – gtrrebel Jan 11 '18 at 13:52
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Another partial result: either $f(x) = 0 $ for all $x$ or $f$ has no real roots.

Proof:

Let $r$ be the largest real root of $f$. Then, by the functional equation, $ 2r^3+r$ is also a root of $f$, and so by the definition of $r$, $$ r \geq 2r^3+r $$ which simplifies to $$ 0 \geq r$$ Similarly, if we let $n$ be the smallest root of $f$, once again by the functional equation $2n^3+n $ is also a root and so $$ n \leq 2n^3+n$$ which simplifies to $$ 0 \leq n$$ So any root of $f$ must be between $0$ and $0$ and hence can only be $0$. So now there are two possibilities: 1) $f$ has no real roots 2) $f$ only has roots at $x=0$. In the second case, $f$ must be of the form $$ f(x) = a x^n $$ for some constant $a$ and non-negative integer $n$. Substituting this in the functional equation yields $$ ax^n a(2x^2)^n = a (2x^3+x)^n$$ $$ a^2 2^n x^{3n} = a (2x^3+x)^n $$ which is only possible if $n=0$ since otherwise the RHS has multiple terms. Hence, we are left with $$ a^2 = a$$ and $$f(x)=a$$ so $a=0$ or $a=1$, but we must take $a=0$ since case 2) is under the assumption that $f$ has a root at $x=0$. This completes the proof.

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Not a solution, but some observations: If $f$ is constant, $f(x) = c$ then $c^2 = c$ therefore $c\in\{0,1\}$.

If $f$ is not constant, but has a root $a$, then (as gtrrebel said) $2a^3+a$ is a root too. If $a = 2a^3+a$, then $a=0$, that means $x$ is a factor of $f$. If $a\neq 2a^3+a$ then we $(x-a)(x-2a^3-a)$ is a factor of $f$. If $a$ is positive we can solve $a=2b^2$ for $b = \sqrt{a/2}$, then $2b^3+b$ is a root too with the same argument.

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  • $\begingroup$ If $a=0$, and $x^k$ is a factor of $f$, then the left-hand side is divisible by $x^{3k}$ and the right-hand side by $x^k$. $\endgroup$ – Empy2 Jan 24 '15 at 21:43
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This is a partial solution as well.
If $f$ has degree $n$, and leading coefficient $c$, then the leading coefficients of the equation are $c^22^n=c2^n$, so $c=1$.
If $f(0)=d$ then $d^2=d$. If $d=0$ then the equation shows that $f(x)$ is divisible by $x^3$; then repeat that and $f(x)$ is divisible by $x^9,x^{27}$ and so on. So $d=1$.
The coefficient of $x^{3n-1}$ is zero on the right-hand side, so it is zero on the left-hand side; so the coefficient of $x^{n-1}$ in $f(x)$ is zero.
$f(x)=x^2+1$ works. (so does f(x)=0, if that is a polynomial)

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    $\begingroup$ $(x^2+1)^k$ works too. $\endgroup$ – Empy2 Jan 24 '15 at 22:07
  • $\begingroup$ that is because if we multiply any two polynomials for which it works, the product will also work $\endgroup$ – Asier Calbet Jan 24 '15 at 22:14

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