(strong) induction proof:$\left(a+\sqrt{b}\right)^n+\left(a-\sqrt{b}\right)^n=2k,a,b,k\in\mathbb N, a>b$

Question

In the very beginning, I'm going to refer to an answer by @BillDubuque: https://math.stackexchange.com/a/48533/721644,

that motivated me to generalize my problem.

Instead of proving $\left(3+\sqrt{5}\right)^n+\left(3-\sqrt{5}\right)^n=2k,k\in\mathbb N$

I wanted to prove:

$\left(a+\sqrt{b}\right)^n+\left(a-\sqrt{b}\right)^n=2k,a,b,k\in\mathbb N, a>b$.

I haven't learned about groups yet. We have only mentioned them once or twice, so I wanted to stick to the given strong induction problem in a more abstract degree.

$(1)$ base case: $\tau(1)$ $$2a\in 2\mathbb N$$ $(2)$ assumption:

Let the statement $$\left(a+\sqrt{b}\right)^n+\left(a-\sqrt{b}\right)^n=2k,k\in\mathbb N$$

hold for some $\{1,\ldots,n\}$.

$(3)$ step: $\tau(n+1)$

\begin{equation}\left(a+\sqrt{b}\right)^{n+1}+\left(a-\sqrt{b}\right)^{n+1}=\left(a+\sqrt{b}\right)^n(a+\sqrt{b})+\left(a-\sqrt{b}\right)^n(a-\sqrt{b})\\a\underbrace{\left(\left(a+\sqrt{b}\right)^n+\left(a-\sqrt{b}\right)^n\right)}_{\in 2\mathbb N}+\sqrt{b}\left(\left(a+\sqrt{b}\right)^n-\left(a-\sqrt{b}\right)^n\right)\end{equation}

\begin{equation}\sqrt{b}\left(\left(a+\sqrt{b}\right)^n-\left(a-\sqrt{b}\right)^n\right)=2b\sum_{i=0}^{n-1}\left(a+\sqrt{b}\right)^{n-1-i}\left(a-\sqrt{b}\right)^i\\=2b\sum_{i=0}^{\left\lceil\frac{n}{2}\right\rceil\\}\underbrace{\left(\left(a+\sqrt{n}\right)^{n-1-2i}+\left(a-\sqrt{n}\right)^{n-1-2i}\right)}_{\in 2\mathbb N}(a^2-b)^i\end{equation}

Therefore, $\left(a+\sqrt{b}\right)^{n+1}\left(a-\sqrt{b}\right)^{n+1}\in 2\mathbb N$ and so is $\left(a+\sqrt{b}\right)^n+\left(a-\sqrt{b}\right)^n$.

Is this correct?

Answer 1

You can prove more using a bit of algebra.

$\mathbb{Z}[\sqrt b]$ is a ring of algebraic integers. Also, $\theta(a+\lambda \sqrt b)=a-\lambda \sqrt b$ where $a$, $\lambda$ $\in \mathbb{Z}$ defines an automorphism $\theta$ of $\mathbb{Z}[\sqrt b]$.

If $a$, $\lambda$ $\in \mathbb{Z}$ then $a+\lambda \sqrt b\in \mathbb{Z}[\sqrt b]$ and hence for $n\in \mathbb{Z_{>0}}$, $(a+\lambda \sqrt b)^n\in \mathbb{Z}[\sqrt b]$ and therefore $(a+\lambda \sqrt b)^n=A+\Lambda \sqrt b$ where $A$, $\Lambda$ $\in \mathbb{Z}$. Applying $\theta$ we also have $(a-\lambda \sqrt b)^n=A-\Lambda \sqrt b$ giving $$(a+\lambda \sqrt b)^n+(a-\lambda \sqrt b)^n=2A\in 2\mathbb{Z}.$$

where $\mathbb{Z}$ can be replaced by $\mathbb{N}$ if either $n$ is even or $a\in \mathbb{N}$.

Answer 2

Here is a different take, which I hope is instructive.

Let $x_n = \left(a+\sqrt{b}\right)^n+\left(a-\sqrt{b}\right)^n$.

Since $a+\sqrt{b}$ and $a-\sqrt{b}$ are roots of $x^2=2ax-(a^2-b)$, we have $x_{n+2}=2ax_{n+1}-(a^2-b)x_n$.

Therefore, if $x_n$ and $x_{n+1}$ are even, then so is $x_{n+2}$.

Thus, the result follows by induction because the base cases are easily checked:

$x_0=2$, $x_1=2a$, $x_2=2(a^2+b)$.