I have tried $f(f(f(f(x))))=f\left(f\left(x^{2}-1\right)\right)$ . Since we know that $f(f(x))=x^{2}-1$, we have $$ \begin{aligned} f\left(f\left(x^{2}-1\right)\right) &=\left(x^{2}-1\right)^{2}-1 \\ &= x^{4}-2 x^{2}+1-1 \\ &=x^{4}-2 x^{2}. \end{aligned} $$But something doesn't feel right, is there any other way to solve this?
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1$\begingroup$ What does not feel right here? The fact that you don't need $f$ itself to find the answer? Finding $f$ is itself part of a whole different field, that of iterative function theory (Read up on "functional square roots"). You were probably expected to solve the question the way you did. $\endgroup$– Sarvesh Ravichandran IyerCommented Mar 8, 2021 at 4:59
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The approach given is correct. Perhaps it would feel more natural to be explicit in our substitution, and write $g(x) = f(f(x))$, so we have $g(x) = x^2-1$, and we are interested in $g(g(x))$.
Then it is easy to see that $g(g(x)) = (x^2-1)^2-1 = x^4-2x^2$.
