Show that $(\sqrt{2} + \sqrt{3})^{2009}$ is rounded to an even number. 
Show that if you round $(\sqrt{2} + \sqrt{3})^{2009}$ to the closest integer you get an even number.

I tried without success to write it in binomial form and to multiply with a conjugate.
edit: Maybe they changed the number for the new course of 2009? im not sure.
Now volfram alfa gives 0 as rest.
http://www.wolframalpha.com/input/?i=round%5B%28sqrt%282%29%2Bsqrt%283%29%29%5E2009%5D+mod+10 
Anyone got any ideas?
Thanks,
 A: This is not a solution, but rather a general way to approach this kind of questions. Maybe you can continue from here:
1) Denote $\alpha=\sqrt{2}+\sqrt{3}$. Find a polynomial over the integers $p(x)$ such that $p(\alpha)=0$, preferably of a small degree. For example, here we have:
$$\alpha^2=2+2\sqrt{6}+3=5+2\sqrt{6}\hspace{5pt}\Rightarrow\hspace{5pt}\alpha^4-10\alpha^2+25=(\alpha^2-5)^2=24$$
So we can choose $p(x)=x^4-10x^2+1$. From the construction it is easy to see that the other roots of $p(x)$ are $\bar{\alpha}=\sqrt{2}-\sqrt{3}$, $-\alpha=-\sqrt{2}-\sqrt{3}$ and $-\bar{\alpha}=-\sqrt{2}+\sqrt{3}$.
2) Consider the linear homogeneous recurrence relation such that $p(x)$ is its characteristic polynomial: $a_n-10a_{n-2}+a_{n-4}=0$. The general solution to this equation is given by $a_n=A_1\alpha^n+A_2(-\alpha)^n+A_3(\bar{\alpha})^n+A_4(-\bar{\alpha})^n$.
Now we need to construct a solution (find some $A_2,A_3,A_4$ and fix $A_1=1$) such that the closest integer to $\sqrt{2}+\sqrt{3})^{2009}$ is either $a_{2009}$ or $a_{2009}-1$.
Having the recurrence relation allows us to prove inductively that for all odd $n$, $a_n$ have the same parity - based only on $a_1$ and $a_3$. 
