If $\left(1+\sqrt2\right)^{2011}=a+b\sqrt{2}$, for integers $a$ and $b$, then what is $\left(1-\sqrt2\right)^{2010}$ expressed using $a$ and $b$? 
Being $ a $ and $ b $ integers such that $\left(1+\sqrt{2}\right)^{2011} =a+b\sqrt{2}, \left(1-\sqrt{2}\right)^{2010}$ equals:
a) $a+2b+(a-b)\sqrt{2}$
b) $a-2b+(a-b)\sqrt{2}$
c) $a+2b+(b-a)\sqrt{2}$
d) $2b-a+(b-a)\sqrt{2}$
e) $a+2b-(a+b)\sqrt{2}$

Solution:

Not going with any of the alternatives

Note that if $(1+\sqrt{2})^n=a_n+b_n \sqrt{2}$ then $(1-\sqrt{2})^n=a_n-b_n \sqrt{2}$.
We have
\begin{eqnarray*}
(1+\sqrt{2})^{2011}=a+b \sqrt{2} \\
(1-\sqrt{2})^{2010}=x+y \sqrt{2}. \\
\end{eqnarray*}
Multiply the second equation by $(1-\sqrt{2})$ & do a bit of algebra ... this gives
\begin{eqnarray*}
x-2y=a\\
x-y=b .\\
\end{eqnarray*}
Should be easy from here ?

 d)

