If $ax+by=7$, . 
If 
  $$ax+by=7$$
  $$ax^2+by^2=49$$
  $$ax^3+by^3=133$$
  $$ax^4+by^4=406$$
  then find the value of
  $$2014(x+y-xy) - 100(a+b)$$

My attempt:
$$ax^2+by^2=49$$
$$ax^2+by^2=(ax+by)^2$$
$$ax^2+by^2=a^2x^2+2abxy+b^2y^2$$
$$ax^2-a^2x^2+by^2-b^2y^2=2abxy$$
$$ax^2(1-a)+by^2(1-b)=2abxy$$

The structure of the equations indicate that $7$, $49$, $133$, $406$, ... is a sequence that is defined as a homogeneous linear difference equation of order $2$ with the characteristic equation $\lambda^2+c_1\lambda +c_0=0$ with the roots $x$ and $y$. Therefore we have 
\begin{eqnarray*}
133 + \;\;49c_1 + \;\;7c_0 &=& 0 \\
406 + 133c_1 + 49c_0 &=& 0
\end{eqnarray*}
with $c_1 = -(x+y)$ and $c_0=xy$. From this, we easily get $x+y$ and $xy.$ Then we can find $x$ and $y$ and use the first two equations to figure out $a$ and $b$.
