<s>[INST] Find $f(2^{2017})$ The function $f(x)$ has only positive $f(x)$. It is known that $f(1)+f(2)=10$, and $f(a+b)=f(a)+f(b) + 2\sqrt{f(a)\cdot f(b)}$. How can I find $f(2^{2017})$?
The second part of the equality resembles $(\sqrt{f(a)}+\sqrt{f(b)})^2$, but I still have no idea what to do with $2^{2017}$.
 [/INST] $f(2) = f(1+1) = f(1) + f(1) + 2 \sqrt{f(1) \cdot f(1)} = 4f(1)$
Together with $f(1) + f(2) = 10$, this gives $f(1) = 2$ and $f(2) = 8$.
Then, $f(n+1) = f(n) + 2\sqrt{2f(n)} + 2 = (\sqrt{f(n)}+\sqrt2)^2$, i.e. $\sqrt{f(n+1)} = \sqrt{f(n)} + \sqrt2$
This tells us that $\sqrt{f}$ is an arithmetic sequence with common difference $\sqrt2$.
With $\sqrt{f(1)} = \sqrt2$ and $\sqrt{f(2)} = 2\sqrt2$, we obtain $\sqrt{f(n)} = n\sqrt2$, i.e. $f(n) = 2n^2$.
In particular, $f(2^{2017}) = 2 \cdot (2^{2017})^2 = 2^{4035}$.
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