Determine all functions $f(n)$ from the positive integers to the positive integers which satisfy the following condition: whenever $a,b,$ and $c$ are positive integers such that $\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$ then $\frac{1}{f(a)} + \frac{1}{f(b)} = \frac{1}{f(c)}$.
Attempt
We want to find the form of $f(n)$. Notice that if $n$ is a positive integer, then the only possible solution to $$ \frac{1}{a} + \frac{1}{b} = \frac{1}{n} $$ is only $a=b=2n$.
So we must only have that $$ f(n) = \frac{f(2n)}{2}$$
If $n=1$ then $2f(1) = f(2)$, if $n=2$ then $2f(2) = f(4)$. $n=3$ then $2f(3) = f(6)$.
$$ f(1) = f(2)/2 $$ $$ f(2) = f(4)/2 $$ $$ f(3) = f(6)/2 $$ $$ f(4) = f(8)/2 $$ $$ f(5) = f(10)/2 $$ $$ \vdots $$
From here we can easily see that $$ f(1) = \frac{f(2)}{2} = \frac{f(4)}{2^{2}} = \frac{f(8)}{2^{3}} = ... $$
So we have $f(2) = 2f(1), f(4) = 4f(1), f(8) = 8f(1), ...$
$$ f(2^{k}) = 2^{k} f(1)...$$
then also $f(3) = f(6)/2 = f(12)/4 = f(24)/8 = …,$ which means
$$ f(3 \times 2^{k}) = 2^{k} f(3) $$