Math Olympiad Algebra Question

These equations are in the form of the general solution of a linear recurrence equation of order $2$ with constant coefficients. The general formula of the sequence is $$ u_n=ax^n+by^n $$ as the solution of a recurrence $$ u_{n+1}=cu_n+du_{n-1}, $$ where $x,y$ are solutions of the characteristic equation $$ 0=λ^2-cλ-d=(λ-x)(λ-y). $$ Knowing two triples from the given four members $(u_1,u_2,u_3,u_4)=(7,49,133,406)=7\cdot(1,7,19,58)$ of the sequence allows to establish two linear equations for $c$ and $d$ \begin{align} 19&=7c+d\\ 58&=19c+7d\\ \text{and consequently}&\\ 75&=30c\\ c&=\frac52\\ d&=19-7c=\frac32 \end{align} so that $$ 2u_{n+2}-5u_{n+1}-3u_n=0. $$ For the roots $x$ and $y$ of the characteristic equation we get $x+y=\frac52$ and $xy=-\frac32$. The expression $a+b=u_0$ is the term of the sequence before the given ones, $$ a+b=\frac13(2u_2-5u_1)=\frac13(98-35)=21 $$ This is sufficient to compute the crazy combination that is requested as answer.

From $ax+by=7$, we have $ax=7-by, by=7-ax$. Noting $$ ax^2+by^2=x\cdot ax+y\cdot by=x(7-by)+y(7-ax)=7(x+y)-(a+b)xy, $$ from $ax^2+by^2=49$, we obtain $$ \tag{$*$} 7(x+y)-(a+b)xy=49. $$ Similarly, $$ ax^2=49-by^2,by^2=49-ax^2, ax^3=133-by^3,by^3=133-ax^3, $$ from which, we have $$ ax^3+by^3=x\cdot ax^2+y\cdot by^2=x(49-by^2)+y(49-ax^2)=49(x+y)-(ax+by)xy $$ and hence $49(x+y)-7xy=133$ or $$ \tag{$**$} 7(x+y)-xy=19. $$ Finally, $$ ax^4+by^4=x\cdot ax^3+y\cdot by^3=x(133-by^3)+y(133-ax^3)=133(x+y)-(ax^2+by^2)xy=133(x+y)-49xy $$ and hence $133(x+y)-49xy=406$ or $$\tag{$***$}\quad\quad\quad 19(x+y)-7xy=58. $$ From $(*), (**), (***)$, it is easy to see $$x+y=\frac{5}{2},xy=-\frac{3}{2},a+b=21 $$ and hence $$ 2014(x+y-xy)-100(a+b)=5956. $$