Determine the value of $ax^5 + by^5$ given that $a$, $b$, $x$, and $y$ satisfy the following system: Given

$\begin{cases} ax + by = 4 \\ ax^2 + by^2 = 8 \\ ax^3 + by^3 = 17 \\ ax^4 + by^4 = 42 \end{cases} $

Determine the value of $ax^5 + by^5$
Do you have any brilliant idea to solve this problem? By looking at the right-hand side, it is actually a sequence of the sum of quadratic numbers. But, of course it is so irrational to say that $ax^5 + by^5 = 42 + 36 = 78$ just by looking at that point. 

You may exploit the idea behind the Berlekamp-Massey algorithm. By denoting as $U_n = a x^n +b y^n $ we have that $\{U_n\}_{n\geq 0}$ is a linear recurrent sequence with a quadratic characteristic polynomial and such that
$$ U_0=a+b,\quad U_1=4,\quad U_2=8,\quad U_3=17,\quad U_4=42 $$
These informations are enough to reconstruct the characteristic polynomial, hence the next term of the sequence. Indeed
$$ \begin{pmatrix}17 & 8 \\ 8 & 4 \end{pmatrix}\begin{pmatrix}u\\ v\end{pmatrix} = \begin{pmatrix}42\\ 17\end{pmatrix} $$
implies that the characteristic polynomial $x^2-ux-v$ is $x^2-8x+\frac{47}{4}$ and
$$ U_5 = 8U_4-\frac{47}{4}U_3 =\color{red}{\frac{545}{4}}. $$
In compact form:
$$ U_5 = (U_4\,U_3)\begin{pmatrix}U_3 & U_2 \\ U_2 & U_1 \end{pmatrix}^{-1}\begin{pmatrix} U_4\\ U_3\end{pmatrix}. $$
