An algebra problem by Phudish Prateepamornkul

The sequence $z_j \,=\, a x^j + b y^j$ will satisfy the recurrence relation $z_{j+2} - (x+y)z_{j+1} + xy z_j \; = \; 0 \hspace{2cm} j \ge 1$ and hence we obtain simultaneous equations $\begin{aligned} 310 - 110(x+y) + 40xy & = 0 \\ 890 - 310(x+y) + 110xy & = 0 \end{aligned}$ Thus we obtain $x+y=5$ and $xy = 6$ . From this it follows that $z_5 = 5\times890 - 6\times310 = \boxed{2590}$ .

If we wanted, we could note that we must have $x=2$ , $y=3$ , and proceed from there to find $a$ and $b$ by solving the equations $\begin{aligned} 2a + 3b & = 40 \\ 4a + 9b & = 110 \end{aligned}$ obtaining $a=5$ and $b=10$ , so that $z_j = 5\times2^j + 10\times3^j$ . This is not necessary to answer to question, though.

Let $c_n=ax^n+by^n$ for $n\in\mathbb{N}$

Let $d_n=c_{n+2}c_{n}-{c_{n+1}}^2$ for $n\in\mathbb{N}$ .

By above, $c_1=40,c_2=110,c_3=310,c_4=890$ and $d_1=300,d_2=1800$ .

Now $\begin{aligned} d_n & = & c_{n+2}c_{n}-{c_{n+1}}^2\\ & = & (ax^{n+2}+by^{n+2})(ax^n+by^n)-(ax^n+by^n)^2\\ & = & (a^2x^{2n+2}+abx^{n+2}y^n + abx^{n}y^{n+2}+b^2y^{2n+2})-(a^2x^{2n+2}+2abx^{n+1} y^{n+1}+b^2y^{2n+2})\\ & = & ab(x-y)^2(xy)^n \end{aligned}$ \ So $d_n$ is a geometric progression, and so $d_3 = 10800$ . $10800 = d_3 = c_3c_5-{c_4}^2 = 310c_5-792100$ and so \ $ax^5+by^5=c_5=\dfrac{10800+792100}{310}=2590$ .