Romanian Mathematical Olympiad

Let $a_0=a$. It is obvious that $a_1=2a$, and, for $n=2$, $$a_2^2-2a{a}_2-8a^2=0,$$ implying $a_2=4a$, for $a_2>0$. Use induction on $n$ to prove that $a_n=2^{n}a$. Assume that $a_k=2^{k}a$, for all $k$, $0\le k\le n$, to prove $a_{n+1}=2^{n+1}a$. We have $$2a{a}_{n+1}+a^2{2}^{n+1}\sum _{k=1}^n{C}_{n+1}^k=a_{n+1}^2,$$ or $$2a{a}_{n+1}+a^2{2}^{n+1}(2^{n+1}-2)=a_{n+1}^2.$$ Since $a_{n+1}$ is positive, we obtain $a_{n+1}=2^{n+1}a$, as needed.