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Find all function $f(x)$ such that $$\large f(x^2 + x + 3)f(3x + 1) = f(6x^3 + 7x^2 + 16x + 3), \forall x \in \mathbb R$$

I have provided my solution but there isn't a strong claim for why $f(x)$ is in the form of $(ax + b)^n$ where $a, b \in \mathbb R, n \ge 1$ (as Luca Bressan suggested).

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  • $\begingroup$ I don’t have an answer, but I have some ideas: Letting $x=0$, we see that: $f(3) f(1) = f(3)$, which implies: $f(3) = 0$ or $f(1)= 1$. $\endgroup$ – Joe Sep 5 '19 at 16:32
  • $\begingroup$ Notice that if $f(x)$ satisfies the property, then also $f(x)^n$ satisfies the property for any $n > 1$. For example, since you have already shown that $f(x) = 2x - 1$ is a solution, then also $f(x) = 4x^2 - 4x + 1$ is a solution. $\endgroup$ – Luca Bressan Sep 5 '19 at 16:35
  • $\begingroup$ I also noticed that $x^2+x+3$ is symmetric about $x=-1/2$, so that so is $f(x^2+x+3)$. I thought maybe factoring that cubic could be helpful, but it doesn’t seem to factor nice. $\endgroup$ – Joe Sep 5 '19 at 16:37
  • $\begingroup$ Is $f$ required to be a polynomial, or at least continuous? If not, there are lots of pathological examples. $\endgroup$ – user125932 Sep 17 '19 at 4:04
  • $\begingroup$ It has to be a polynomial. $\endgroup$ – Lê Thành Đạt Sep 17 '19 at 5:30
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I think the following can help. Define:$$a=x^2+x+{5\over 2}\\ b=6x+1$$and $g(x)=f(x+0.5)$, therefore$$g(a)g(b)=g(2ab)$$

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    $\begingroup$ I suppose you mean $b = 3x + \frac 1 2$. $\endgroup$ – Luca Bressan Sep 5 '19 at 16:46
  • $\begingroup$ Strictly speaking this argument only shows that $g(a) g(b) = g(2 a b)$ when at least one of $a, b$ is $\geq \frac{9}{4}$ (the minimum value of $x^2 + x + \frac{5}{2}$). $\endgroup$ – Travis Willse Sep 5 '19 at 17:55
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It can be seen that $f(x) = (ax + b)^n$ $(a, b \in \mathbb R, n \ge 1)$. We have that $$[a(x^2 + x + 3) + b][a(3x + 1) + b] = a(6x^3 + 7x^2 + 16x + 3) + b, \forall x \in \mathbb R$$

$$\iff a^2(3x^3 + 4x^2 + 10x + 3) + ab(x^2 + 4x + 4) + b^2$$

$$ = a(6x^3 + 7x^2 + 16x + 3) + b, \forall x \in \mathbb R$$

$$\iff 3a^2x^3 + (4a^2 + ab)x^2 + (10a^2 + 4ab)x + (3a^2 + 4ab + b^2)$$

$$ = 6ax^3 + 7ax^2 + 16ax + (3a + b), \forall x \in \mathbb R$$

$$\implies \left\{ \begin{align} 3a^2 &= 6a\\ 4a^2 + ab &= 7a\\ 10a^2 + 4ab &= 16a\\ 3a^2 + 4ab + b^2 &= 3a + b\end{align} \right., \forall x \in \mathbb R$$

$$\iff (a, b) \in \{(0, 0), (0, 1), (2, -1)\} \iff f(x) \in \{0, 1, (2x - 1)^n\}, \forall x \in \mathbb R, n \ge 1$$

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    $\begingroup$ "It can be seen that". How can this be seen? (isn't this really what the question asks about?) $\endgroup$ – Winther Sep 5 '19 at 16:15
  • $\begingroup$ That is what I ask in the question. I can't really demonstrate how that happens. $\endgroup$ – Lê Thành Đạt Sep 5 '19 at 16:16

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