One side of the implication is obvious (put n=2). Suppose $f(a^2) = 2f(a)$. Notice that if $f(a^{2^k}) = 2^kf(a)$, it implies $f(a^{2^{k+1}}) = 2f(a^{2^k}) = 2^{k+1}f(a)$, basically the induction step that will show $f(a^{2^k}) = 2^kf(a)$ for naturals k. (*) Also, notice $f(a^n) \geq nf(a)$ by $f(a^2 * a^{n-2}) \ge 2f(a) + f(a^{n-2})$ (very obvious). Less obvious is that $2^k f(a) = f(a^{2^k}) \ge f(a^{2^k - 1}) + f(a)$ so that if $f(a^k) = kf(a)$ for k, then also for all naturals $j \le k$. But from (*) k can be arbitrarily large.

Singular wrote: Notice that if $f(a^{2^k}) = 2^kf(a)$, it implies $f(a^{2^{k+1}}) = 2f(a^{2^k}) = 2^{k+1}f(a)$ Why?

I think it doesn't necessarily imply that..

if $f(a^{2^k})=2^kf(a)$ then $f(a^{2^{k+1}})=f(\left(a^{2^k}\right)^2)=2\cdot f(a^{2^k})=2^{k+1}f(a)$ i dont see the problem, but singular didnt prove both ways only one ...

Singular wrote: One side of the implication is obvious (put n=2). sorry singular i overread this

peeta wrote: if $f(a^{2^k})=2^kf(a)$ then $f(a^{2^{k+1}})=f(\left(a^{2^k}\right)^2)=2\cdot f(a^{2^k})=2^{k+1}f(a)$ i dont see the problem, but singular didnt prove both ways only one ... Did I misunderstand the question? I think $f(x^2) = 2f(x)$ works only for $x=a$, not $x = a^{2^k}$

it works for any $a\in\mathbb Z \backslash {0}$

OK. I thought it mean if we pick $a$ and $f(a^2) = 2f(a)$ then $f(a^n) = nf(a)$