Folklore

The function $f(x)$ is defined on the positive half-line and takes only positive values. It is known that $f(1)+f(2)=10$ and $f(a+b)=f(a)+f(b)+2 \sqrt{f(a) f(b)}$ for any $a$ and $b$. Find $f\left(2^{2011}\right)$.
From the condition of the problem, it follows that $f(2 a)=f(a+a)=4 f(a)$. In particular, $f(2)=4 f(1)$, so $5 f(1)=10$, which means $f(1)=2$.

The sequence $f(1), f(2), f\left(2^{2}\right), \ldots$ is a geometric progression with a common ratio of 4. The number $f\left(2^{2011}\right)$, as the 2012th term of this progression, is $2 \cdot 4^{2011}=2^{4023}$.

## Answer

$2^{4023}$.