Show that $(1+\sqrt{2})^n+(1-\sqrt{2})^n$ is the nearest integer to $(1+\sqrt{2})^n$

Question

For $n\geq 1$, show (without expanding brackets) that $(1+\sqrt{2})^n+(1-\sqrt{2})^n$ is an integer, and that moreover it is the nearest integer to $(1+\sqrt{2})^n$.

This is a (non-assessed) problem in an example sheet I am working on. The example sheet is for an algebraic number theory course so I imagine that we are expected to use some tools from algebraic number theory to attack this.

I've tried some things but I can't make any progress in even showing that it's an integer (though it's obvious if you expand the brackets). I wonder if we're expected to do something a bit creative like showing that it's the norm/trace of some algebraic integer or the discriminant of some number field, which would suffice but sounds a bit far-fetched. Taking a more elementary approach I realise we have $a^n+b^n$ where $a+b=2,ab=-1$ and we could consider odd and even $n$ separately but I can't do anything with this. I don't want to completely waste this problem, so instead of looking at the solution I'd appreciate being given a small hint to get me started.

Answer 1

HINT: The number $(1+\sqrt{2})^n+(1-\sqrt{2})^n\in\Bbb{Z}[\sqrt{2}]$ is unchanged by substituting $\sqrt{2}$ by $-\sqrt{2}$.

---

My original (over)complete answer:

To show that it's an integer...

...first note that it's an element of $\Bbb{Z}[\sqrt{2}]$, so it is of the form $a+b\sqrt{2}$ for some $a,b\in\Bbb{Z}$. It's invariant under the automorphism of $\Bbb{Z}[\sqrt{2}]$ that maps $\sqrt{2}$ to $-\sqrt{2}$. This means $b=0$ and so it's an integer.

To show that it is the nearest integer to $(1+\sqrt{2})^n$...

...it suffices to show that $|(1-\sqrt{2})^n|<\frac{1}{2}$.

Answer 2

$(1\pm \sqrt 2)$ are the roots of the quadratic $x^2-2x-1$.

It follows that $a_n=(1+\sqrt 2)^n+(1-\sqrt 2)^n$ satisfies the recursion $$a_{n+2}=2a_n+a_{n-1}$$

Since $a_0=2,a_1=2$ we see that all the $a_n$ are integers.

The other claim follows since $$n≥1 \implies |(1-\sqrt2)^n|<.5$$

(which in turn follows from the fact that $|1-\sqrt 2|\approx .41$)

Answer 3

Partial answer.

$a:=1+√2$, $b=1-√2$;

Induction.

$n=1 √.$

Hypothesis:

$a^n+b^n$ is an integer.

Step $n+1:$

$a^{n+1}+b^{n+1}= aa^n+bb^n=$

$(1+√2)a^n+(1-√2)b^n=$

$(a^n+b^n) +√2(a^n-b^n);$

The first summand is an integer by hypothesis.

The second summand:

Recall:

$a^n-b^n=$

$\small{(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2...+b^{n-1})}$.

Let $n$ be even:

Grouping together, first and last term, second and second last term, etc.:

$√2(a-b)×$

$[(a^{n-1}+b^{n-1}) +(a^{n-2}b +b^{n-2}a)+$

$(a^{n-3}b^2+b^{n-3}a^2)+ ..... $

$a^{n-(k+1)}b^k+a^kb^{(n-(k+1)}....]$,

where $k=0,1,.....(n/2-1)$.

1) $a^{n-1}+b^{n-1}$, is integer (hypothesis)

2) $aba^{n-3}+abb^{n-3}=$

$(ab)^1(a^{n-3}+b^{n-3})$,

where $(ab)$, $(a^{n-3}+b^{n-3})$(hypothesis) are integers.

$k-$th term:

$(ab)^{k}(a^{n-(2k+1)}+b^{n-(2k+1)})$.

First factor is an integer, as is the second by hypothesis.

A similar reasoning for n is odd. Grouping together as before woks, but for one term of the form

$a^{(n-1)/2}b^{(n-1)/2}$(No partner).

What can yo say about this term?