Show that $\forall n\in\Bbb{N}, (3+\sqrt 7)^n+(3-\sqrt 7)^n\in\Bbb{Z}$ and that $\forall n\in\Bbb{N}, (2+\sqrt 2)^n+(2-\sqrt 2)^n\in\Bbb{Z}$ I got this problem which I encountered during a limit of sequence calculation:
Show that $\forall n\in\Bbb{N}, (3+\sqrt 7)^n+(3-\sqrt 7)^n\in\Bbb{Z}$
And that
$\forall n\in\Bbb{N}, (2+\sqrt 2)^n+(2-\sqrt 2)^n\in\Bbb{Z}$
I tried induction but I failed to show it.
Thanks on any help.
Note: I don't know advanced number theory.
 A: Way 1:  Imagine expanding using the Binomial Theorem, and adding. There is nice cancellation of the terms that involve odd powers of $\sqrt{7}$.
Way 2:  Our sum is invariant under the mapping that sends numbers of the form $a+b\sqrt{7}$, where $a$ and $b$ are integers, to $a-b\sqrt{7}$. But it is not hard to see that only integers are invariant under that mapping.
Way 3:  We can also use a recurrence.
Note that $3+\sqrt{7}$ and $3-\sqrt{7}$ are roots of the equation $x^2-6x+2=0$. Let $a_n=(3+\sqrt{7})^n$ and $b_n=(3-\sqrt{7})^n$. It is easy to verify that $a_{n+2}=6a_{n+1}-2a_n$ and $b_{n+2}=6b_{n+1}-2b_n$.
Let $c_n=(3+\sqrt{7})^n+(3-\sqrt{7})^n$. Then by linearity we have
$$c_{n+2}=6c_{n+1}-2c_n.\tag{1}$$
Note that $c_0=2$ and $c_1=6$, both integers. It follows by induction using (1) that $c_n$ is an integer for all $n$.  
Way 4: For brevity write our sum as $\alpha^n +\beta^n$. Note that
$$\alpha^{n+1}+\beta^{n+1}=(\alpha^n +\beta^n)(\alpha +\beta)-\alpha\beta(\alpha^{n-1}+\beta^{n-1}).\tag{1}$$ 
Since $\alpha+\beta$ and $\alpha\beta$ are integers, using (1) we can show that
if $\alpha^n+\beta^n$ and $\alpha^{n-1}+\beta^{n-1}$ are integers, then so is
$\alpha^{n+1}+\beta^{n+1}$. This does the induction step. 
A: If you know a bit of abstract algebra, there is nothing to compute. (And this is one of those examples which show that abstraction doesn't make things complicated, but in fact easier.)
Consider the field extension $\mathbb{Q}(\sqrt{7})$ of $\mathbb{Q}$. Its automorphism group is generated by $\sqrt{7} \mapsto -\sqrt{7}$. Since $(3+\sqrt{7})^n + (3-\sqrt{7})^n \in \mathbb{Q}(\sqrt{7})$ is obviously invariant under this transformation, it lies in $\mathbb{Q}$. It is also integral over $\mathbb{Z}$ (since $3,\sqrt{7}$ are integral). Since $\mathbb{Z}$ is integrally closed, the element actually lies in $\mathbb{Z}$.
The same argument works for $(2+\sqrt{2})^n+(2-\sqrt{2})^n$ and tons of other examples. For example, $(1+i)^n + (1-i)^n \in \mathbb{Z}$ by considering the field of Gaussian rationals $\mathbb{Q}(i)$.
Edit: As explained by André Nicolas  in  this answer, the argument gets even simpler by working directly with the ring $\mathbb{Z}[\sqrt{7}]$.
A: We have
$$\begin{align}(3+\sqrt 7)^n+(3-\sqrt 7)^n&=\sum_{k=0}^{n}\binom{n}{k}3^{n-k}(\sqrt 7)^k+\sum_{k=0}^{n}\binom{n}{k}3^{n-k}(-\sqrt 7)^k\\&=\sum_{k=0}^{n}\binom{n}{k}3^{n-k}(\sqrt 7)^k(1+(-1)^k).\end{align}$$Here, note that if $k$ is odd, then 
$$(\sqrt 7)^k(1+(-1)^k)=0$$
and that if $k$ is even, then $$(\sqrt 7)^k(1+(-1)^k)\in\mathbb Z.$$
A: If $a$ and $b$ are any integers (positive or negative), then $(a+\sqrt b)^n + (a-\sqrt b)^n$ is an integer. This is because if you expand it out using the binomial expansion, the terms $(\sqrt b)^m$ from $(a+\sqrt b)^n$ cancel out the terms $(-\sqrt b)^m$ from $(a-\sqrt b)^n$ for odd $m$.
A: Hint:
In the binomial expansion of
$$(3+\sqrt 7)^n+(3-\sqrt 7)^n$$
we have terms such as
$$3^n+3^{n-1}\sqrt 7{n\choose 1}+3^{n-2}\sqrt 7^2{n\choose 2}+\dots$$
What happens when two such binomials are expanded, particularly when one of them involves a negative?
Alternative hint:
What happens in the following multiplication:
$$((2+\sqrt2)^n+(2-\sqrt2)^n)((2+\sqrt2)+(2-\sqrt2))$$
(This one may lend itself to induction...)
A: It's a great idea to look at $\mathbb{Z}[\sqrt{7}]$.
But if you haven't gotten to algebraic number theory in your studies just yet, let's try basic algebra. I shall use $n$ here to refer to 0 or a positive integer.
What's the formula for $(3 - \sqrt{7})^n$? Define the recurrence relations $a_n = 6a_{n - 1} - 2a_{n - 2}$, with $a_0 = 1, a_1 = 3$ and $b_n = 6b_{n - 1} - 2b_{n - 2}$ with $b_1 = 1, b_2 = 6$. Then, $(3 - \sqrt{7})^n = a_n - b_n \sqrt{7}$ for $n > 0$. Verify this before going further. Verify also that $a_n$ and $b_n$ are integers.
Now, what's the formula for $(3 + \sqrt{7})^n$? It turns out to be very similar, and it even employs the same recurrence relations defined earlier. Thus, $(3 + \sqrt{7})^n = a_n + b_n \sqrt{7}$.
Putting it all together: $(3 - \sqrt{7})^n + (3 + \sqrt{7})^n = a_n - b_n \sqrt{7} + a_n + b_n \sqrt{7} = 2a_n$.
