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This is a Mathematica-based question derived directly from this math question.

Given that f[x] is monotonically increasing and f[f[x]] == x^2 + 2, find f[11], where x and f[x] are in the integers.

I tried applying inverse functions (and found the inverse of the compounded functions is $-\sqrt{x-2}$), and all obvious applications of Solve without luck. Note that f[f[3]]=11... if perhaps that is of use.

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    $\begingroup$ Possibly related to mathematica.stackexchange.com/questions/227863/… and mathematica.stackexchange.com/questions/817/… but for integers instead of the reals. Unfortunately the Carleman matrix method won't work here due to the nature of x^2+2 - it's an even function so the matrix is singular. $\endgroup$ – flinty Sep 22 at 11:57
  • $\begingroup$ f[x_] := If[x > 0, -x^2, -x + 2] will work if you sacrifice 'monotonically increasing' and restrict the range to x > 0. The first application of y = f[x] will 'save` or 'memorize' the value of x^2 by sticking it in the negative numbers so that the second application knows if y is negative then it just has to add two. Of course, this feels like cheating, but it leads me to suspect it's impossible to construct a function that has all these properties simultaneously and works on all the integers. If the domain is finite, there may be a permutation which works though. $\endgroup$ – flinty Sep 22 at 21:59
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I came across this answer here https://math.stackexchange.com/a/3599697/775158. It's miraculous that it works, at least I think so. The resulting half-iterate $h$ is not perfect as $\forall x\in\mathbb{Z} : h(x)$ is not necessarily an integer. But it is true that $\forall x\in\mathbb{Z^+}: h(h(x))\in\mathbb{Z^+}$ and $h$ is monotone over the positive reals.

g[n_] := Nest[Function[{x}, Evaluate[Sqrt[#[(x^2 + 2)] - 2]]] &, 
  Function[{x}, Abs[x]^Sqrt[2]], n]

(* we get very good convergence for n > 3 *)
h = g[4];

(** g[4] is this thing
Function[{x}, Sqrt[-2 + 
  Sqrt[-2 + 
   Sqrt[-2 + 
    Sqrt[-2 + Abs[2 + (2 + (2 + (2 + x^2)^2)^2)^2]^Sqrt[2]]]]]]
**) 

N[h[h[3]]]
(* result 11. *)

So we have a good approximation for the half-iterate like this: $$ h(x)\approx\sqrt{\sqrt{\sqrt{\sqrt{\left(\left| \left(\left(\left(x^2+2\right)^2+2\right)^2+2\right)^2+2\right| ^{\sqrt{2}}\right)^{\sqrt{2}}-2}-2}-2}-2} $$ I also looked for an integer $x$ such that $h(x)$ is close to an integer and I found $h(36)\approx159.001$. We also have $h(11)\approx 30.014$.

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    $\begingroup$ Gee.... that's a lot of work and an awkward function. Definitely worth $+1$, but I'll keep the question open just in case there's a "more perfect" (and integer) solution. Thanks! $\endgroup$ – David G. Stork Sep 22 at 21:26
  • $\begingroup$ Yes functional equations are very weird and the literature on them is all over the place. There doesn't seem to be a one-size-fits-all solution to these kinds of problems out there. I tried my pytorch neural network solver from my question I linked in the comments. There's something about the evenness / symmetry of some functions that makes the convergence very poor. Whereas an odd function like a cubic would probably have a strange half iterate but one that could be easily found using the Carleman matrices or the Newtonian method as I mentioned earlier. $\endgroup$ – flinty Sep 22 at 21:33
  • $\begingroup$ Take a look here: math.stackexchange.com/questions/3835531/… $\endgroup$ – David G. Stork Sep 22 at 21:55

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