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Prove without using complex numbers that $f:\mathbb{R}→\mathbb{R}$ does not exist such that $f(f(x))=x^2+4x+3$. Note that for example if $g(g(x))=x^2-2$ then $g(g(x))$ has two real fixed points -1 and 2; also $g(g(g(g(x))))$ has 4 fixed points -1 and 2 and $\frac{\sqrt5-1}{2}$ and $-\frac{\sqrt5+1}{2}$. Now we can see with some accuracy that there is no such function $g$ that $g(g(x))=x^2-2$. But $f(f(x))=x^2+4x+3$ and $f(f(f(f(x))))$ (and similar functions) does not have real fixed points. If these problems are solved, the questions like the following link can be resolved decisively: On the functional square root of $x^2+1$ I have a solution using complex numbers but I tried to find a preliminary solution to it but I did not succeed. Thanks

mofidy
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    What have you tried? – Thomas Andrews Aug 27 '19 at 17:56
  • Related: https://math.stackexchange.com/questions/3633/on-the-functional-square-root-of-x21 – Jack D'Aurizio Aug 27 '19 at 18:17
  • Solve as much as possible using high school math,Please – mofidy Aug 27 '19 at 18:25
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    @mofidy That's not how MSE works. We are not here to do your homework for you. Do you BTW ask for the non-existence of any function or is $f$ supposed to be continuous or even analytic? – amsmath Aug 27 '19 at 18:29
  • @JackD'Aurizio Your function is only defined on $[-4,0]$.... – amsmath Aug 27 '19 at 19:26
  • Put $g(x) := f(x-2)+2$. Then the functional equation is equivalent to $g(g(x)) = x^2+1$. Due to https://math.stackexchange.com/questions/3633/on-the-functional-square-root-of-x21 this equation does have a solution on $\mathbb R$. – amsmath Aug 27 '19 at 20:54
  • @JackD'Aurizio Assume that $f\in C(\mathbb R)$ satisfies $f(f(x)) = (x+2)^2-4$. Then clearly $f$ is injective on both $I_- = (-\infty,-2)$ and $I_+ = [-2,\infty)$. Assume that there is $x < -2$ with $f(x) < -2$. If $f$ is increasing on $I_-$, then $f(x-1) < f(x) < -2$ and so $f(f(x-1)) < f(f(x))$ which is a contradiction as $x^2+4x$ is decreasing on $I_-$. If $f$ is decreasing on $I_-$, choose $a>0$ such that $x+a < -2$ and conclude $f(x+a) < f(x) < -2$ and hence $f(f(x+a)) > f(f(x))$, which again is a contradiction. This implies that $f(I_-)\subset I_+$.... – amsmath Aug 27 '19 at 21:58
  • Now, if $f$ is increasing on $I_-$ it follows even that $f(I_-)\subset [-2,f(-2)]$. On the other hand, $f$ must be decreasing on $I_+$ (since otherwise $f$ and thus also $f\circ f$ would be injective on $\mathbb R$). Hence, $f$ is bounded from above which then also holds for $f\circ f$ -- a contradiction. Therefore, $f$ must be decreasing on $I_-$ and increasing on $I_+$ with values in $I_+$, i.e., $f(x)\ge -2$ for all $x\in\mathbb R$. But $f(f(-2)) = -4$, a contradiction! Such a function can therefore not exist. – amsmath Aug 27 '19 at 21:59
  • @amsmath: all right, so we have analytic solutions in a neighbourhood of $-2$, but no continuous and global solution. On the other hand, we should have a plethora of global and discontinuous solutions. Am I wrong? – Jack D'Aurizio Aug 27 '19 at 23:56
  • @JackD'Aurizio I don't know. I'm not an expert on this. Seems to be a difficult topic. Just tried to get a contradiction for the original equation. It didn't work out, but on the way (the stuff above) I found that your equation cannot have a solution. – amsmath Aug 28 '19 at 00:00

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