a property in some real number $\alpha=2+\sqrt{2}$ and $\beta=2-\sqrt{2}$ let $\alpha=2+\sqrt{2}$ and $\beta=2-\sqrt{2}$ 
i want to show for $n\in\Bbb N$
$$ \alpha^{n}+\beta^{n}\in \Bbb N$$ and $$\lim\limits_{n\rightarrow\infty}\alpha^{n}+\beta^{n}=[\alpha^{n}]+1$$
thanks for any help.
 A: Let $f_n = \alpha^n + \beta^n$. Consider, for some $\mu$ and $\nu$
$$
      \mu f_{n} + \nu f_{n-1} = \left(\mu \alpha + \nu \right) \alpha^{n-1} + \left(\mu \beta + \nu\right) \beta^{n-1}
$$
Let's find $\mu$ and $\nu$ such that:
$$
   \mu \alpha + \nu  =\alpha^2, \quad \mu \beta + \nu = \beta^2 \quad \implies \quad \mu= \alpha+ \beta= 4, \quad \nu = -\alpha \beta = -2
$$
Hence, $f_n$ satisfies recurrence equation $f_{n+2}  = 4 f_n - 2 f_{n-1}$. Since initial conditions $f_0=2$, $f_1=4$ are integers, $f_n \in \mathbb{N}$ follows, since $f_n >0 $ by construction.
Furthermore, since $0<\beta = \frac{2}{\alpha}<1$, $\beta^n$ rapidly decreases as $n$ grows, yielding the second assertion.
$$
    \alpha^n + \beta^n = [ \alpha^n ] + \{ \alpha^n \} + \beta^n \to [\alpha]^n + 1
$$
A: Hint $\ $ Use induction and the following universal recurrence
$$\rm a^{n+2} + b^{n+2}\, =\, (a+b)\, (a^{n+1}+b^{n+1}) - ab\,(a^n+b^n)\quad [here\ \ \ a+b=4,\ ab = 2]$$
A: For the first part, try writing out the binomial expansions of $\alpha^n$ and $\beta^n$.
A: HINTS: To show that $\alpha^n+\beta^n\in\Bbb N$, use the binomial theorem. To show that $$\lim_{n\rightarrow\infty}\left(\alpha^{n}+\beta^{n}\right)=\left\lfloor\alpha^{n}\right\rfloor+1\;,$$ use the fact that $|\beta|<1$.
