<s>[INST] $1+\sqrt{2}$ seems special: its powers quickly approach integers $ \alpha = 1+\sqrt{2}$ has some interesting properties. One of which is that if you look at powers
\begin{align*}
\alpha^2 &= 3+2\sqrt{2} \approx 5.8284 \\
\alpha^3 & \approx 14.0710 \\
\alpha^4 & \approx 33.9705 \\
& \vdots \\
\alpha^{10} & \approx 6725.9998 \\
\alpha^{11} & \approx 16238.0000 \\
\vdots & 
\end{align*}
(values truncated, not rounded) then the sequence $\{ \alpha^n \}$ is "almost" an unbounded integer sequence. What is going on here? Is this property secondary to $\alpha$ being a unit in $\mathbb{Z}[\sqrt{2}]$? Can $\alpha$ be interpreted as a type of approximate eigenvalue? Does this behavior have to do with the continued fraction representation?
Other numbers such as $1+\sqrt{3}$ have a similar property, but they don't appear to "become integer" as quickly. 
 [/INST] The key fact here turns out to be that the conjugate (in this case in $\Bbb Q[\sqrt{2}]$) of $1 + \sqrt{2}$, namely, $1 - \sqrt{2}$, has absolute value less than $1$.
If we expand 
$$(1 + \sqrt{2})^n + (1 - \sqrt{2})^n$$
and apply the Binomial Theorem to both of the two powers, we find all of the nonintegral terms cancel, so $(1 + \sqrt{2})^n + (1 - \sqrt{2})^n$ is an integer for all $n$. But since $|1 - \sqrt{2}| < 1$, for large $n$ the term $(1 - \sqrt{2})^n$ tends to $0$ and so $$(1 + \sqrt{2})^n + (1 - \sqrt{2})^n \approx (1 + \sqrt{2})^n ,$$ and in particular $(1 + \sqrt{2})^n$ is close to an integer.
We can identify the sequence of integers that occurs, too: $(1 \pm \sqrt{2})$ are the roots of the polynomial $r^2 - 2 r - 1$, so $(1 + \sqrt{2})^n + (1 - \sqrt{2})^n$ satisfies the recurrence relation $$a_{n + 2} = 2 a_{n + 1} + a_n .$$ Evaluating at, say, $n = 0, 1$ gives $a_0 = a_1 = 2$, and these values and the recursion relation are enough to identify the sequence,
$$2, 2, 6, 14, 34, \ldots .$$
This sequence is OEIS A002203, where its members are called the companion Pell numbers. (Its entries are exactly twice those of OEIS A001333, $1, 1, 3, 7, 17, \ldots$, the numerators of the continued fraction convergents of $\sqrt{2}$.) Since the initial values are integers, as are the coefficients in the recursion relation, this gives an alternative way of seeing that $(1 + \sqrt{2})^n + (1 - \sqrt{2})^n$ is an integer for all $n$.
Finally, working backward, we can indeed interpret $1 + \sqrt{2}$ (and its conjugate) as eigenvalues of a suitable matrix: The companion matrix associated to the polynomial $r^2 - 2 r - 1$ above is $\pmatrix{0&1\\1&2}$, and its eigenvalues are $1 \pm \sqrt{2}$. Then, a straightforward induction shows that 
$$\pmatrix{0&1\\1&2}^n \pmatrix{2&2\\2&6} = \pmatrix{a_n&a_{n + 1}\\a_{n + 1}&a_{n+ 2}} .$$
All of these phenomena, by the way are in close analogy to the more familiar case of the Golden Ratio, $\phi := \frac{1}{2}(1 + \sqrt{5})$. (You might work these details out to check that you understand the above arguments.) Indeed, $1 + \sqrt{2}$ has a geometric interpretation analogous to that of the Golden Ratio, and by analogy it is sometimes called the Silver Ratio; in turn these are the first two (nontrivial) members of the infinite family $\frac{n + \sqrt{n^2 + 4}}{2}$, $n = 0, 1, 2, \ldots$, of metallic means, all of which (except for the trivial case $n = 0$) have the key property re the conjugate.

Like you've observed, $1 + \sqrt{3}$ also has the key property, since $|1 - \sqrt{3}| < 1$. But $|1 - \sqrt{3}| > |1 - \sqrt{2}|$, so $|1 - \sqrt{3}|^n \to 0$ more slowly than $|1 - \sqrt{2}|^n \to 0$.
</s>