jmc

For $n=1,2,3,4,5,$ define $s_n=a{x}^n+b{y}^n.$ We are given the values of $s_1,s_2,s_3,$ and $s_4,$ and want to compute $s_5.$

We find a relationship between the terms $s_n.$ Notice that $$\begin{array}{rl}(x+y)(a{x}^n+b{y}^n)& =a{x}^{n+1}+bx{y}^n+a{x}^{n}y+b{y}^{n+1}\\ & =(a{x}^{n+1}+b{y}^{n+1})+xy(a{x}^{n-1}+b{y}^{n-1}).\end{array}$$In other words, $(x+y)s_n=s_{n+1}+xy{s}_{n-1}$ for all $n.$ Therefore, taking $n=2$ and $n=3,$ we get $$\begin{array}{rl}7(x+y)& =16+3xy\\ 16(x+y)& =42+7xy.\end{array}$$Solving this system of equations for $x+y$ and $xy$ gives $x+y=-14$ and $x=-38.$ Thus, taking $n=4,$ we get $$42(x+y)=s_5+16xy,$$so $$s_5=42(-14)-16(-38)=\boxed{20}.$$