<s>[INST] Functional Equation $f\big(f(x)\big)=3x$ over natural nubers for strictly increasing $f$ 
If $f$ is a strictly increasing function from the naturals to the naturals, and $f\big(f(x)\big)=3x$, what are all values of $f(2012)$?

I have only proven that $f(3x)=3f(x)$ but that gets nowhere :(
 [/INST] Proposition. $f(1) = 2$, $f(2) = 3$.
Proof. $f(1)$ cannot be $1$ otherwise we would have 
$$
3 = f(f(1)) = f(1) = 1
$$
So $f(1) > 1$ and since $f$ is strictly increasing, $f(x) > x$ for each $x$.
Being $3 = f(f(1)) > f(1)$, the only remaining possibility is $f(1) = 2$.
Finally, $f(2) = f(f(1)) = 3$. $\blacksquare$
Proposition. $f(3^n x) = 3^n f(x)$.
Proof. This is a direct consequence of the relation
$$
f(3x) = f(f(f(x))) = 3f(x)\; \blacksquare
$$ 
Proposition. If $3^n < x \leq 2\cdot 3^n$ then $f(x) = 3^n + x$.
Proof. It's sufficient to note that there are exactly $3^n$ numbers between $3^n$ and $2\cdot 3^n$, and exactly $3^n$ numbers between 
$$
f(3^n) = 3^nf(1)= 2\cdot 3^n
$$ 
and 
$$
f(2\cdot 3^n) = 3^nf(2) = 3^{n +1} \; \blacksquare
$$
Proposition. If $2\cdot 3^n < x \leq 3^{n + 1}$ then $f(x) = 3x - 3^{n + 1}$.
Proof. Since $3^n < x - 3^n \leq 2\cdot 3^n$, the previous proposition implies $f(x -3^n) = x$ and therefore
$$
f(x) = f(f(x - 3^n)) = 3x - 3^{n + 1} \; \blacksquare
$$
Being $2\cdot 3^6 < 2012 \leq 3^7$, from last proposition we can conclude
$$
f(2012) = 3\cdot 2012 - 3^7 = 3849
$$ 
</s>