How to come up with proofs of these results? Or, are these results true in the first place?

Question

Let $x_n$ and $y_n$ be integer sequences determined by $$x_n + y_n \sqrt{2} = (1+\sqrt{2})^n \ \ \ \mbox{ for } \ n= 1, 2, 3, \ldots. $$

Then how to show that

(a) $x_{n+1} = y_{n+1} + y_n$, $\ \ \ \ \ $ $y_{n+1} = x_n + y_n$.

(b) $y_{n+1} = y_{n+1}^2 + y_n^2$.

(c) $y_{n+2} = 2y_{n+1} + y_n$. How to derive a similar formula for $x_{n+2}$?

(d) $x_{2n+1} = 2x_{n+1} x_n + (-1)^{n+1}$.

(e) How to derive general formulas for $x_n$ and $y_n$?

I was hoping to use induction or to be able to relate two terms of either sequences in some way but haven't had much success either way.

I wonder which area of mathematics this result belongs to!

And, I would be really grateful for a reference to some good book that deals with such matters.

Answer 1

The numbers defined by $x_{n} + y_{n} \, \sqrt{2} = (1+\sqrt{2})^{n}$ are called Pell and Pell-Lucas numbers.

As stated by the proposer question B should be $y_{2n+1} = y_{n+1}^{2} + y_{n}^{2}$ as will be shown.

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1) From $x_{n} + y_{n} \sqrt{2} = (1+\sqrt{2})^{n}$ it is seen that \begin{align} x_{n+1} + y_{n+1} \sqrt{2} &= (1+\sqrt{2})^{n+1} = (1+\sqrt{2}) \cdot (x_{n} + y_{n} \sqrt{2}) \\ &= (x_{n} + 2 y_{n}) + (x_{n} + y_{n}) \sqrt{2} \end{align} which upon equating the coefficients the results become \begin{align} x_{n+1} &= x_{n} + 2 y_{n} \\ y_{n+1} &= x_{n} + y_{n}. \end{align} Notice that $x_{n}$ can also be seen in the form $x_{n+1} = y_{n+1} + y_{n}$.

2) difference equations: Let $n \to n+1$ in $y_{n+1} = x_{n} + y_{n}$ to obtain $y_{n+2} = y_{n+1} + x_{n+1}$. Now using the relation for $x_{n+1}$ this becomes $y_{n+2} = 2 y_{n+1} + y_{n}$. For the $x_{n}$ equation consider the following: \begin{align} x_{n+2} &= y_{n+2} + y_{n+1} = x_{n+1} + 2 y_{n+1} \\ &= x_{n+1} + 2 x_{n} + 2 y_{n} = x_{n+1} + 2 x_{n} + x_{n+1} - x_{n} \\ &= 2 x_{n+1} + x_{n}. \end{align}

3) explicit values: It can be determined that the same form remains when $\sqrt{2} \to - \sqrt{2}$ which leads to \begin{align} x_{n} + y_{n} \sqrt{2} &= (1+\sqrt{2})^{n} = a^{n} \\ x_{n} - y_{n} \sqrt{2} &= (1-\sqrt{2})^{n} = b^{n}. \end{align} Adding and subtracting these relations it is evident that \begin{align} x_{n} &= \frac{a^{n} + b^{n}}{2} \\ y_{n} &= \frac{a^{n} - b^{n}}{2 \sqrt{2}}. \end{align}

4) value of $y_{n+1}^{2} + y_{n}^{2}$: \begin{align} y_{n+1}^{2} + y_{n}^{2} &= \frac{1}{8} \left[ (a^{n+1} - b^{n+1})^{2} + (a^{n} - b^{n})^{2} \right] \\ &= \frac{1}{8} \left[ 2(2+\sqrt{2}) \, a^{2n} + 2 (2 - \sqrt{2}) \, b^{2n} \right] \\ &= \frac{a^{2n+1} - b^{2n+1}}{2 \sqrt{2}} \\ &= y_{2n+1}. \end{align}

5) $x_{2n+1}$ equation. \begin{align} 2 x_{n+1} x_{n} &= \frac{1}{2} \, (a^{n+1} + b^{n+1})(a^{n} + b^{n}) \\ &= \frac{1}{2} ( a^{2n+1} + b^{2n+1}) + \frac{1}{2} (ab)^{n} (a+b) \\ &= x_{2n+1} + (-1)^{n} \end{align} which can be seen in the form $x_{2n+1} = 2 x_{n+1} x_{n} - (-1)^{n}$.

Answer 2

For example, and using an inductive argument:

$$x_{n+1}+y_{n+1}\sqrt2=(1+\sqrt2)^{n+1}=(1+\sqrt2)(1+\sqrt2)^n=(1+\sqrt2)(x_n+y_n\sqrt2)\implies$$

$$\implies x_{n+1}+y_{n+1}\sqrt2=(x_n+2y_n)+(x_n+y_n)\sqrt2$$

Thus, we have

$$\begin{cases}y_{n+1}=x_n+y_n\\{}\\x_{n+1}=x_n+2y_n\end{cases}\implies x_{n+1}=(y_{n+1}-y_n)+2y_n=y_{n+1}+y_n$$

Try now to take it from here.

Answer 3

The unit group of the ring of integers $\mathbb{Z}[\sqrt{2}]$ is given by $\{ \pm(1+\sqrt{2})^n\mid n\in \mathbb{N}\}$, i.e., $1+\sqrt{2}$ is a fundamental unit of the real quadratic number field $\mathbb{Q}(\sqrt{2})$. Note that $(1+\sqrt{2})(\sqrt{2}-1)=1$, hence $(1+\sqrt{2})^n(\sqrt{2}-1)^n=1$, so that all $(1+\sqrt{2})^n$ are invertible in the ring $\mathbb{Z}[\sqrt{2}]$.

We have $(1+\sqrt{2})^n=x_n+y_n\sqrt{2}$, with $(a)$ and $(c)$. This can be proved by induction. However, $(b)$ is not true, because already $y_{n+1}^2>y_{n+1}$, and $x_n,y_n\ge 1$.

Answer 4

In fact $$\eqalign{ x_n &= \dfrac{(1 + \sqrt{2})^n + (1 - \sqrt{2})^n}{2}\cr &= \sum_{j=0}^{\lfloor n/2\rfloor} {n \choose 2j}\; 2^j\cr y_n &= \dfrac{(1 + \sqrt{2})^n - (1 - \sqrt{2})^n}{2\sqrt{2}}\cr &= \sum_{j=0}^{\lfloor (n-1)/2 \rfloor} {n \choose 2j+1} \; 2^j }$$

The recursions can be derived from

$$\eqalign{x_{n+1} + y_{n+1} \sqrt{2} &= (1 + \sqrt{2}) (x_{n} + y_{n} \sqrt{2})\cr &= (x_n + 2 y_n) + (x_n + y_n) \sqrt{2}}$$ so $x_{n+1} = x_n + 2 y_n$ and $y_{n+1} = x_n + y_n$. But note that you can replace $\sqrt{2}$ by $-\sqrt{2}$: $$\eqalign{x_{n+1} - y_{n+1} \sqrt{2} &= (x_n + 2 y_n) - (x_n + y_n) \sqrt{2}\cr &= (1 - \sqrt{2})(x_n - y_n \sqrt{2})}$$ so you also have $x_n - y_n \sqrt{2} = (1 - \sqrt{2})^n$. Putting this and $x_n + y_n \sqrt{2} = (1+\sqrt{2})^n $ together, we can solve for $x_n$ and $y_n$.

For more (and there's much more), you might look at OEIS sequences A001333 and A000129.