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. 
 A: 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:
$$ U_5 = (U_4\,U_3)\begin{pmatrix}U_3 & U_2 \\ U_2 & U_1 \end{pmatrix}^{-1}\begin{pmatrix} U_4\\ U_3\end{pmatrix}. $$
A: By taking resultants for the polynomials we obtain "linear relations", namely
$$
x=8-y,\quad y= \frac{47(17a + 16)}{392},\quad b= \frac{ - 47a + 96}{47}
$$
Then the system is equivalent to one quadratic equation in $a$, namely
$$
799a^2 - 1632a + 16=0.
$$
Then we obtain
$$
ax^5+by^5=\frac{545}{4}.
$$
A: You have
$$ (x+y)(ax^2+by^2) = ax^3+by^3+xy(ax+by) \implies 8(x+y) = 17+4xy  $$
$$ (x+y)(ax^3+by^3) = ax^4+by^4 + xy(ax^2+by^2) \implies 17(x+y) = 42 + 8xy $$
It follows that 
$$ x+y = 8 $$
$$ xy = \frac{47}{4} $$
Also, from the same relations
$$ ax^5 + by^5 = (x+y)(ax^4+by^4) - xy(ax^3+by^3) = 42(x+y)-17xy = \frac{545}{4} $$
A: I don't know if this is the fastest way, and it doesn't seem brilliant, but what the heck.
We can take the second equation to produce two equations out of it:
$$ax^3+bxy^2=8x\\ax^2y+by^3=8y$$
summing the two gives
$$ax^3+by^3+xy(ax+by) = 8(x+y)\\
17 + 4xy=8(x+y)$$
Using the same process on the third equation gives the equation
$$42 + 8xy = 17(x+y)$$
so we now have two equations for two variables, $x, y$, and we can calculate $x,y$, and from that, $a,b$.

As I said, I don't think this is the simplest way, but it seems to get the job done.
