So I've been going through some of my textbooks to keep my mathematics knowledge up on point, and I can across this problem which I'm not quite sure how to solve. I've tried a few methods (I'll put my attempts below), but I just can't find a good way to solve this.

Let's define $f(x) = \sqrt {4x^2-4x^4}$. Consider the equation $f(f(f(...(f(x))...))) = x$, where there are a total of $2020$ functions $f$ being applied to $x$. And let's say a total of $A$ real numbers satisfy this equation. Find the value of of $A$ (Mod $1000$).

I first though of trying to take the amount of functions that it would take to get a certain number back to itself, and check whether that would be divisible by $2020$, because if it is, then that means it will satisfy the condition. I tried to do this for a while, but I just couldn't make sure I was $100%$ accurate and I wasn't sure when to end my search either. And considering the problem wants me to give the answer mod $1000$, I figured that brute forcing it like this would take too long. So then I tried to look for a pattern by looking at the first few numbers that would work, but I just couldn't find any detectable pattern or similar properties of those numbers. And that brings me here, where I am asking for some help to find an efficient method for this problem or possibly a formula for this kind of problem. (And I want to learn from this so a decent explanation or guide would be nice)

Okay so I've got an answer of 576 from somewhat brute forcing this (using above method), but I really want to see if anyone can spot a pattern and confirm this, or use a formula to get an answer and explain their logic.

  • $\begingroup$ You suggest that you have found some solutions...which ones have you found? $\endgroup$
    – lulu
    Jun 17, 2020 at 22:32
  • $\begingroup$ @lulu, so I'll elaborate a bit. I went through all of the numbers that would let the functions repeat themselves, and tried to find the ones that would let the problem repeat itself. For example, how if a number were to revert itself after 2 operations, doing 1010 of those operations would get you the same result at the end. $\endgroup$
    – Matt Klien
    Jun 17, 2020 at 22:36
  • $\begingroup$ Another thing I've noticed is that with each new function, the radicals of the previous function would either be taken to the second or forth power, essentially canceling them out. So f(f(x)) would be equal to $\sqrt(4(\sqrt(4x^2-4x^4))^2-4((4x^2-4x^4))^4)=x$ which is equal to $\sqrt((16x^2-16x^4)-4(4x^2-4x^4)(4x^2-4x^4))$. I'm not sure how to use that fact though. $\endgroup$
    – Matt Klien
    Jun 17, 2020 at 22:39
  • $\begingroup$ @lulu More specifically, I found $x=0$, $x=\sqrt{3}/2$, $1/2 \sqrt{5/2 - \sqrt{5}/2}$, and $1/2 \sqrt{5/2 + \sqrt{5}/2}$ to work for just $f(f(x))=x$. $\endgroup$
    – Matt Klien
    Jun 17, 2020 at 22:41
  • $\begingroup$ For $f(x)$, I found $x=0$, $x=\sqrt{3}/2$ $\endgroup$
    – Matt Klien
    Jun 17, 2020 at 22:47

1 Answer 1


$\mathcal { Hint:} $

$$x=\sin(t)$$ $$f(x)=\sin(2t)$$

After $k$ iterations

$$f(f(...(f(x))...))=\sin(2^kt)$$ Also, since $x \geq0$ , we need to check solutions in only $0 \leq t\lt \pi$.

See if this helps.


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