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 $2^{2017}$.
Hint: Observe we have \begin{align} f(2^{n+1}) = f(2^n + 2^n) = (\sqrt{f(2^n)}+\sqrt{f(2^n)})^2 = 2^2 f(2^n) \end{align}
$f(2) = f(1+1) = f(1) + f(1) + 2 \sqrt{f(1) \cdot f(1)} = 4f(1)$
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, $f(2^{2017}) = 2 \cdot (2^{2017})^2 = 2^{4035}$.
$f(2a) = f(a+a) = f(a) + f(a) + 2\sqrt{f(a)f(a)} = 4f(a)$
By induction $f(2^k)=4f(2^{k-1}) = 4^2f(2^{k-1}) = 4^kf(1)$.
$f(1) + f(2) = f(1) + 4f(1) = 5f(1) = 10$.
So $f(1) = 2$.
So $f(2^k) = 4^kf(1) = 2*4^k$
$f(2^{2017} = 2*4^{2017}=2^{4035}$
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What what be interesting is what other values are.
$f(3) = f(1) + f(2) + 2\sqrt{f(1)f(2)} = 2 + 8 + 2\sqrt {16} = 18$
$..... = 3^2*2?????$
Does $f(k) = 2k^2$?????
$f(1) = 2*1^2$ and $f(2) = 8 = 2*2^2$ and
if $f(k) = 2k^2$ then $f(1+k) = f(1) + f(k) + 2\sqrt{f(1)f(k)}$
$= 2 + 2k^2 + 2\sqrt{2*2k^2} = 2+2k^2 + 4k = 2(k^2 + 2k + 1) = 2(k+1)^2$.
So that holds for integers.
