How to deal with a case where one answer says "there is no solution possible" and another answer shows one?

Question

In this question the OP asks for a solution for the problem:

Find all $f(x)$ satisfying $ f(f(x)) = x^2-2$.

One answer squarely states, that no solution exists (and even gives a reference to a jstor-accepted article which states this as a general theorem on this type of problem), and in another answer someone else (in this case I) happen to show one solution.

So - being an amateur - I first assumed that something with my solution was wrong (but had no idea about what this could be). After that, I suspected that the general theorem might be only for a certain sub-class of that type of problems, but I cannot find a key for that in the article; on the contrary, the authors emphasize that indeed no solution of whatever form can exist.

Now that question/answers are of 2013, so of nearly 2 years ago and the two statements still stay parallel in the same question, stating opposite "truths".

How can I / we resolve that problem? Put a comment below of each of the controversy statements? Put a decisive comment below the question?

[Updated question]

Shall we leave the contradicting answers as they are (which, in my view is a very uneasy situation for any casual reader without own deeper insight) or should some comment be made, possibly at each of the contradictory answers, or possibly the authors be pinged to add some explanations into their answers in the view of the concurring statements about the existence of a solution?

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[added some information]

Here is a bit from the article:

THEOREM 1. Let $P$ be a polynomial of degree 2 defined on the entire complex plane C. Then P has no iterative roots of any order whatever; i.e., for any integer $r \ge 2$, there exists no function $f$ whatever mapping C into itself such that $f°^r= P$.

It must be emphasized that the phrase "no function f whatever" is to be taken literally and does not mean merely "no entire function" or "no continuous function," etc. From this it is to be expected that none of the usual methods of analysis or topology play a role in what follows. This is the case. The proof of Theorem 1 is, in essence, purely combinatorial.

I assume this is the part which leads to the introduction in the answer which refers to that article:

There is no such f

Answer 1

There are fixed points where the derivative is neither $1$ nor $-1.$ As a result, there is (maybe) a real analytic solution on $(-\infty, -1),$ another on $(-1,2),$ definitely a third on $(2,\infty).$ It will not be possible to join these into a single real analytic function. However, a letter from Jean Ecalle may apply here (I cannot be quite sure), saying that the three solutions can be joined into a single $C^\infty$ solution. Such a thing does apply for $f(f(x)) = \sin x,$ so that the real solution I constructed for MO is $C^\infty$ and piecewise $C^\omega$

As I mentioned, the derivatives at the fixed points do not have unit magnitude. As a result, on the three regions, solutions can be found using the easier Schröder equation.

This is discussed in adequate detail in KCG, which I bought used at a reasonable price. Give me a few minutes, I will find some applicable pages. PAGE 354, Theorem 8.6.1. Before I forget, I have always been uneasy about functions with two fixpoints and the interval between them; the methods I know go on one side of a fixpoint only. Maybe I should say, about my first paragraph, that I suspect there is a real analytic solution on $(-1,2).$ I recall there were some difficulties with negative derivative at the fixpoint, from an early MO question about $g(g(x)) = \cos x.$

Note: I found where I saved Milnor's book on Complex Dynamics. It appears that the linearization theorem of Koenigs guarantees real analytic fractional iterates of a real analytic function around a fixpoint $x_0,$ as long as either $0 < f'(x_0) < 1$ or $1 < f'(x_0).$ Alright, in Daniel S. Alexander, A History of Complex Dynamics pages 46-47, section 3.8, we find that the solution of the Schröder equation in KCG is due to Koenigs, and he proved analyticity. Anyway, the inability to extend a fractional iterate to all of $\mathbb C$ need not prevent pretty good solutions on all or part of $\mathbb R.$

Alright, it turns out that the Koenigs stuff does almost everything, but that a local minimum or maximum, despite not being a fixpoint, causes trouble anyway. For $x^2-2,$ there will be a real analytic half iterate on $(0, \infty),$ but we will need to use $g(|x|)$ to get the whole real line; a bit unpredictable about smoothness at $x=0.$ The problem for comparison is $h(h(x)) = x^2,$ with $h(x) = |x|^{\sqrt 2}.$ Our $x^2 - 2$ might do better, it might do the same.

Answer 2

After the recent edit by san, the "There is no such $f$" answer has become self-contained. Reiterating it here:

Let $F_n$ denote the set of fixed points of $f^{\circ n}$. Then we have the trivial facts that $f(F_n)\subseteq F_n$ and that $n\mid m$ implies $F_n\subseteq F_m$. One computes $$F_2=\{-1,2\},\qquad F_4=\left\{-1,2,\frac{-1+\sqrt5}2,\frac{-1-\sqrt5}2\right\}.$$ As $f\left(f\left(\frac{-1\pm\sqrt 5}{2}\right)\right)=\frac{-1\mp\sqrt 5}2\notin F_2$, we conclude that $f$ maps $S=F_4\setminus F_2$ to itself. As $|S|=2$, $f|_S$ must have a fixed point or permute $S$ in which case $f^2|_S$ is the identity, i.e., $S\subseteq F_2$, contradiction.

Your answer in the other direction on the other hand, I fail to find conclusive ...