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
 A: The key point is that $\left(1+\sqrt2\right)^{2011}=a+b\sqrt{2}$ implies
$\left(1-\sqrt2\right)^{2011}=a-b\sqrt{2}$.
Therefore,
$$
\left(1-\sqrt2\right)^{2010}
=\frac{\left(1-\sqrt2\right)^{2011}}{1-\sqrt2}
=\frac{\left(1-\sqrt2\right)^{2011}(1+\sqrt2)}{(1-\sqrt2)(1+\sqrt2)}
=-(a-b\sqrt2)(1+\sqrt2)
= \cdots
$$
A: 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)

A: Let $x= \left(1-\sqrt{2}\right)^{2010}$ 
Then $$x(a+b\sqrt{2}) = (1-2)^{2010}\left(1+\sqrt{2}\right)=1+\sqrt{2}$$
So $$x= {1+\sqrt{2}\over a+b\sqrt{2}}  =  {(1+\sqrt{2})( a-b\sqrt{2})\over a^2-2b^2}$$
Now $\left(1-\sqrt{2}\right)^{2011} = a-b\sqrt{2}$ so $$a^2-2b^2 = (1-\sqrt{2})^{2011} \cdot \left(1+\sqrt{2}\right)^{2011} =-1$$
So $$x=(1+\sqrt{2})( -a+b\sqrt{2}) =...$$
