Problem - 3645
Let real numbers $a, b, c, d$ satisfy
$$
\left\{
\begin{array}{ccl}
ax+by&=3\\
ax^2+by^2&=7\\
ax^3+by^3&=16\\
ax^4 + by^4 &=42
\end{array}
\right.
$$
Find $ax^5+by^5$.
Assume $x$ and $y$ are two roots of $t^2 + mx + n=0$. It is corresponding to sequence $F_{n+2} + mF_{n+1}+nF_n=0, (n\ge 3)$ which is equivalent to $$F_{n+2}=-mF_{n+1}-nF_n$$
Then we have
$$
\left\{
\begin{array}{ccl}
16 &= - 7m - 3n\\
42 &= -16m - 7n
\end{array}
\right.
\implies
(m,n)=(14,-38)
$$
It follows that $$F_5=-mF_4 - nF_3 = -14\times 42 +38\times 16= \boxed{20}$$
(The required knowledge of this solution is explained in the book Competition Algebra .)