2020 CIME II Problems/Problem 9

Solution

We can start by finding the number of solution for smaller repeptitions of $f$ . Notice that we can solve $f(f(x))=f(x)$ by applying the functional inverse $f^{-1}$ to both sides as you would to solve any equation: $f^{-1}(f(f(x)))=f^{-1}(f(x))\Longrightarrow f(x)=|x|$ (We put the absolute value bars because we know that taking the inverse of $f$ of both sides involves taking the square root of both sides, and $\sqrt{t^2}=|t|$ ). From here, it is easy to see that this equation has $4$ solutions at $\pm1$ and $\pm2$ . We can also try for $f(f(f(x)))=f(f(x))$ (we will solve more methodically here): $((x^2-2)^2-2)^2-2=(x^2-2)^2-2$ $(x^2-2)^2-2)^2 = (x^2-2)^2$ $(x^2-2)^2-2=|x^2-2|$ $(x^2-2)^2-2=\pm(x^2-2)$ $(x^2-2)^2=x^2 \textup{ OR } (x^2-2)^2=-x^2+4$ The first equation yeilds $4$ results, and the second equation yields $2$ results for a total of $6$ results. It appeats that $\underbrace{f(f\cdots f}_{n\textup{ times}}(x))=\underbrace{f(f\cdots f}_{n-1\textup{ times}}(x))$ bas $2n$ real solutions, giving a total of $6060$ apparent solutions for the original equation. This makes logical sense considering that $f$ is an even polynomial with 2 roots. For a more formal proof, we consider $F_n(x)=\underbrace{f(f\cdots f}_{n\textup{ times}}(x))$ . We are asked to find the number of solutions of the equation in the form $F_n(x)=F_{n-1}(x)$ . Following from how we colved the first simple case, $f^{-1}(F_n(x))=f^{-1}(F_{n-1}(x)) = F_{n-2}(x)=|F_{n-1}(x)|$

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2020 CIME II Problems/Problem 9

Solution