functional equation:$f(f(x))=6x-f(x)$

Question

Prove that there exists a unique function $f:R^{+}\rightarrow R^{+}$ $$f(f(x))=6x-f(x)$$

My Try

Define $a_{k+1}=f(a_k)$ then we have the recursive relation $$a_{k+2}+a_{k+1}-6a_k=0$$ whose characteristic equation is $$x^{k+2}+x^{k+1}-6x^k=0$$ $$x^2+x-6=0 \Rightarrow x=-3 ,x=2$$ ie $$a_k=c_1 {(-3)}^k+c_2{(2)}^k$$ .As $x>0 \Rightarrow a_0>0\Rightarrow 2c_2>3c_1$

i am stuck now as i have not been able to find $c_1,c_2$

Answer 1

After OP's work: For $a_{k+2}+a_{k+1}-6a_k=0$, take $a_k=t^k$ to get $t_{1,2}=2,-3$. Choose only positive root to write $a_k=C 2^k \implies a_0=C=x$ (by assumption), next $a_1=C. 2=2x$. By assumption $f(x)=a_1.$ So you get $f(x)=2x.$

Note: here $$a_0=x, a_1=f(x),a_2=ff(x), a_3=fff(x),....,a_k=f^{k}(x).$$