<s>[INST] Using induction, show ${(1+\sqrt{2})}^{2n}+{(1-\sqrt{2})}^{2n}$ is an even integer. I'm having serious difficulties with that task, so it should be nice, if there is someone that can help!
The task says:

Prove that the number 
  $${(1+\sqrt{2})}^{2n}+{(1-\sqrt{2})}^{2n}$$ is an even integer for all $n\in\mathbb{N}$

 [/INST] If $a_n=pr^n+qs^n$, with $r\ne s$, the sequence $a_n$ satisfies the recurrence relation
$$
a_0=p+q,\qquad a_1=pr+qs,\qquad a_{n+2}-(r+s)a_{n+1}+rsa_n
$$
Indeed,
\begin{align}
a_{n+2}-(r+s)a_{n+1}+rsa_n
&=pr^{n+2}+qs^{n+2}-(r+s)(pr^{n+1}+qs^{n+1})+rs(pr^n+qs^n) \\
&=p(r^{n+2}-r^{n+2}-sr^{n+1}+sr^{n+1}) \\
&\qquad+q(s^{n+2}-rs^{n+1}-s^{n+2}+rs^{n+1})\\
&=0
\end{align}
In our case $p=q=1$, $r=(1+\sqrt{2})^2=3+2\sqrt{2}$, $s=(1-\sqrt{2})^2=3-2\sqrt{2}$; thus $r+s=6$ and $rs=1$.
Therefore the recurrence relation is
$$
a_0=2,\qquad a_1=6, \qquad a_{n+2}-6a_{n+1}+a_n=0
$$
Since $a_0$ and $a_1$ are even (base of the induction), if we assume $a_k$ even for every $k$ with $0\le k<n+2$, then
$$
a_{n+2}=6a_{n+1}-a_n
$$
is even too.

A similar strategy is considering $b_n=(1+\sqrt{2})^n+(1-\sqrt{2})^n$. The recurrence relation is
$$
b_0=2,\qquad b_1=2,\qquad b_{n+2}-2b_{n+1}-b_n=0
$$
We want to see that $b_{2n}$ is even; the statement is true for $n=0$. Suppose $b_{2n}$ is even; then
$$
b_{2n+2}=2b_{n+1}+b_n
$$
is even as well.
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