Suppose $f(ax)=(f(x))^2-1$. Expanding in power series, we get $a=1+\sqrt{5}$ or $1-\sqrt{5}$. We take positive $a$. If $f\neq{\rm const}$ then $f'(0)\neq0$ - perhaps, it can be any non-zero number. After that, we can uniquely define coefficients in power series $f^{(n)}(0)/n!$ step by step using differentiation of the functional equation $$ f(0)=f(0)^2-1\ \Rightarrow\ f(0)=\frac{1\pm\sqrt{5}}2;\ f'(0)f(0)=af'(0)\ \Rightarrow\ f(0)=a/2;\ \ 2f''(0)f(0)+2(f'(0))^2=a^2f''(0)\ \Rightarrow\ f''(0)=\frac{2(f'(0))^2}{a^2-a};\ \ .... $$ Can $f$ be expressed in terms of some known functions?
It seems that $f$ is analytic everywhere. Due to the Leibnitz formula, we have $$ a^nf^{(n)}(0)=\sum_{k=0}^n\binom{n}{k}f^{(n-k)}(0)f^{(k)}(0), $$ which leads to $$ f^{(n)}(0)=\frac{\sum_{k=1}^{n-1}\binom{n}{k}f^{(n-k)}(0)f^{(k)}(0)}{a^n-a}. $$ It is true that $|f''(0)|\leq|f'(0)|^2=:C^2$, see above. Suppose that we already proved $|f^{(k)}(0)|\leq C^k$, $1\leq k\leq n-1$. Then $$ |f^{(n)}(0)|\leq \frac{C^n2^n}{(\sqrt{5})^{n}}\leq C^n. $$ Hence, the power series converges everywhere.
(The functional equation is somewhat similar to $\cos 2x=2\cos^2x-1$.
If we denote $g(x)=f(x)/f(0)$ then $$ g'(x)=g'(0)g(a^{-1}x)g(a^{-2}x)g(a^{-3}x).... $$
Also, the polynomial $$ P_n(x)=f(a^nf^{-1}(x))=P\circ...\circ P(x),\ \ \ P(x)=x^2-1 $$ is some analog of Chebyshev polynomial $T_{2^n}(x)$.)
from Hot Weekly Questions - Mathematics Stack Exchange
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