Balkan 2012 shortlist

Solution 1. Multiplying $b_n=2a_{n-1}+4b_{n-1}$ by $t\in \mathbb{R}$ and adding $a_n=9a_{n-1}-2b_{n-1}$ we have $$a_n+t{b}_n=(9+2t)a_{n-1}+(-2+4t)b_{n-1}.$$ Selecting $t$ such that $9+2t=(-2+4t)/t$, that is, $t=-1/2$ or $t=-2$; this becomes $$a_n+t{b}_n=(9+2t)(a_{n-1}+t{b}_{n-1})$$ and by induction we have $$a_n+t{b}_n=(9+2t{)}^n(a_0+t{b}_0)=(9+2t{)}^n(1+t).$$ The last equality becomes $$2b_n-a_n=5^n\text{ and }2a_n-b_n=8^n$$ for $t=-2$ and $t=-1/2$, respectively. Adding these we obtain $c_n=a_n+b_n=8^n+5^n$. Assume that for some indices $k<r<m$ we have $c_r^2=c_k{c}_m$. Then $$(8^r+5^r{)}^2=(8^k+5^k)(8^m+5^m)$$ leads to a contradiction as $8^m+5^m$ has at least one prime factor that does not divide $8^r+5^r$ for any $r<m$ by Zsigmondy Theorem.

Solution 2. It can be easily verified that $c_0=2$, $c_1=13$ and $c_n=13c_{n-1}-40c_{n-2}$ for $n\ge 2$. Since the roots of ${\lambda }^2-13\lambda +40=0$ are 5 and 8, we obtain $c_n=5^n+8^n$ for all $n\ge 0$. We will now show that $$c_k{c}_{2r-k-1}<c_r^2<c_k{c}_{2r-k},$$ which, together with the monotonicity of $(c_n{)}_{n=1}^{\infty }$, will imply that no such $(k,r,m)$ exists. The right hand side inequality is equivalent to $$2\cdot 5^r\cdot 8^r<5^k{8}^{2r-k}+5^{2r-k}{8}^k,$$ which follows from the AM-GM inequality. The left hand side inequality follows from $$c_r^2>8^{2r}>2\cdot 8^k\cdot 2\cdot 8^{2r-k-1}\ge c_k{c}_{2r-k-1}$$ as $8^n<c_n\le 2\cdot 8^n$ for all $n\ge 0$.