$c_n+4b_n=a_n+5b_n=19a_{n-1}+18b_{n-1}=18c_{n-1}+a_{n-1}\Rightarrow c_n=18c_{n-1}+a_{n-1}-4b_n=18c_{n-1}-7a_{n-1}-16b_{n-1}=18c_{n-1}-5c_{n-1}-(2a_{n-1}+11b_{n-1})=18c_{n-1}-5c_{n-1}-40c_{n-2}=13c_{n-1}-40c_{n-2}.$ We will prove by induction that $c_n=5^n+8^n.$ We find that $c_1=5+8$ and $c_2=5^2+8^2$, so $P(1)$ and $P(2)$ are right, assume for $0<x<n+1$ that $P(x)$ is true, then $c_{n+1}=13c_n-40c_{n-1}=13*5^n+13*8^n-40*5^{n-1}-40*8^{n-1}=5^{n+1}+8^{n+1} \Rightarrow P(n+1)$ is true $\Rightarrow$ $c_n=5^n+8^n$ for every $n.$ Now assume there exist $k,r,m$ such that $c_r^2=c_mc_k \Rightarrow (5^r+8^r)^2=(5^m+8^m)(5^k+8^k) \Rightarrow m$ or $k$ is bigger than $r$, wlog $m>r$, then by Zsigmondy's Theorem $5^m+8^m$ is divided by a prime that doesn't divide $5^r+8^r \Rightarrow$ there are no $ k,r,m$ that satisfy $c_r^2=c_mc_k.$