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?

[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

  • 1
    $\begingroup$ I will check, but the usual theorem is that there is no entire solution to a half-iteration problem with a target function that is real on the real axis, while there is one in an open set containing the real axis; this is the case with Helmut Kneser's half iterate of $e^x.$ $\endgroup$
    – Will Jagy
    Jun 17, 2015 at 22:09
  • 1
    $\begingroup$ If I were you, I'd continuously study the topic so that I could further improve my answer. If your already fluent in this topic, then work towards improving how you present your answer. Eventually others will either find an error or agree with your solution. $\endgroup$
    – k170
    Jun 17, 2015 at 22:11
  • 2
    $\begingroup$ Your solution has the domain $[2,\infty)$, as you say in your answer. I assume that the theorems apply to the domain $\mathbb R$. $\endgroup$ Jun 17, 2015 at 22:13
  • $\begingroup$ @k170 - yes, surely. (I'm fiddling with this subject since 2007 and mostly in the "tetration-forum"). My question here has more the focus: should we do something in the referred question to resolve the apparent discrepancy between the answers about the existence of a solution for some casual reader who might come across that MSE-question and does not know what he/she should believe? $\endgroup$ Jun 17, 2015 at 23:47
  • 5
    $\begingroup$ Perhaps we have discovered that long sought-for contradiction in Mathematics. This will make it so much easier to prove the Riemann Hypothesis (and its negation, too). $\endgroup$ Jun 18, 2015 at 0:45
  • 8
    $\begingroup$ BTW wouldn't this be more appropriate question on the main than on meta? Making a post asking for clarification of some point in an older posts (in this case it would be clarification of an apparent discrepancy between the two answers is perfectly fine use of the site. It was discussed, for example, here: Clarify an old answer. (And also in some posts that are linked to that one.) $\endgroup$ Jun 18, 2015 at 5:14

2 Answers 2


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.

  • $\begingroup$ Yes, Will, I'll be happy to see that - happy for me. I think from this we should also transfer the information into the OP's question and into comments to the answers to inform some casual reader, what the situation actually is concerning the question of the OP... (It's tagged as contest-question I think) $\endgroup$ Jun 17, 2015 at 23:44
  • 8
    $\begingroup$ probably Meta is not the place to answer the mathematical problem $\endgroup$
    – GEdgar
    Jun 19, 2015 at 15:35
  • $\begingroup$ So what is the real answer to the question? Is Gottfried's solution correct? $\endgroup$
    – user21820
    Jun 23, 2015 at 13:44
  • 4
    $\begingroup$ @user21820, yes, he restricts to an infinite interval on the real line. The article quoted in the other answer discusses extending to all of $\mathbb C$ and says that cannot be done. Both answers are correct. $\endgroup$
    – Will Jagy
    Jun 23, 2015 at 17:36
  • $\begingroup$ @WillJagy: Okay thanks! $\endgroup$
    – user21820
    Jun 24, 2015 at 6:25

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 ...


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .