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 $2^{2017}$.

$f(2a) = f(a+a) = f(a) + f(a) + 2\sqrt{f(a)f(a)} = 4f(a)$
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.
