If $(3+\sqrt{8})^{2023} = a + b\sqrt{8}$ where $a$ and $b$ are integers, then $a$ is the product of two numbers which differ by 2.

Question

For clarity, I'm using the short-hand $\alpha = 3 + \sqrt{8}$ and $\beta = 3 - \sqrt{8}$.

$\boxed{\alpha + \beta = 6\text{ and }\alpha \beta = 1}$.

$\text{Define } (b_n)_{n \geq 1} := \frac{\alpha^{2n-1}+\beta^{2n-1}}{2}$ (this will always give us the rational part of the number $(3+\sqrt{8})^{2n-1}$).

$b_1 = 3 \text{; } b_2 = 99 \text{; } b_3 = 3363$; ...

My task is then equivalent to showing $b_{1012} = m(m+2)$ for some natural $m$, or, equivalently, to show that $b_{1012} + 1$ is a perfect square.

What I've managed to show/notice:

• $b_1 = 1 \cdot 3$, $b_2 = 9 \cdot 11$, $b_3 = 57 \cdot 59$, ... hence I conjecture they all satisfy the condition I'm asked to prove for $b_{1012}$.
• $b_{n+1} = 34b_{n} - b_{n-1} \text{, } \forall n \geq 2$;
• $\lim_{n \to \infty} \frac{b_{n+1}}{b_{n}} = 17 + 6\sqrt{8} = \alpha^2$;
• $\forall n \geq 1 \text{, } b_{n} = (A)b_{2} + (B)b_1$ for whole $A$ and $B$.
• A cool fact that pops out from my calculations: $17 + 6\sqrt{8} \approx 34-\frac{1}{34}$.

I would love some further insight into this problem and/or a full solution will also be appreciated.

Source: Moldavian Math Republican Olympiad Grade 12.

Answer 1

$\alpha =3+2\sqrt2=(\sqrt2+1)^2=\gamma^2$
$\beta =3-2\sqrt2=(\sqrt2-1)^2=\delta^2$ $$b_n+1 = \frac{\alpha^{2n-1}+\beta^{2n-1}+2}{2} = \left(\frac{\gamma^{2n-1}+\delta^{2n-1}}{\sqrt 2}\right)^2 $$ because $2n-1$ is odd, the value inside the parentheses is whole.

Answer 2

Although a bit more involved, here is a solution using linear recurrence relations.

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Deriving Recurrences for $\pmb{a_n}$ and $\pmb{b_n}$

Define $a_n,b_n\in\mathbb{Z}$ as $a_n+b_n\sqrt8=\left(3+\sqrt8\right)^n$. Then $$ \begin{align} a_n+b_n\sqrt8 &=\left(a_{n-1}+b_{n-1}\sqrt8\right)\left(3+\sqrt8\right)\tag{1a}\\ &=(3a_{n-1}+8b_{n-1})+(a_{n-1}+3b_{n-1})\sqrt8\tag{1b} \end{align} $$ Equating the coefficients of $1$ and $\sqrt8$ in $(1)$, we get $$ \begin{align} a_n&=3a_{n-1}+8b_{n-1}\tag{2a}\\ b_n&=\phantom{3}a_{n-1}+3b_{n-1}\tag{2b} \end{align} $$ Then $$ \begin{align} a_n&=3a_{n-1}+8a_{n-2}+24b_{n-2}\tag{3a}\\ &=6a_{n-1}-a_{n-2}\tag{3b} \end{align} $$ Explanation:
$\text{(3a):}$ substitute $n\mapsto n-1$ in $\text{(2b)}$ to get $b_{n-1}=a_{n-2}+3b_{n-2}$
$\phantom{\text{(3a):}}$ then use for $b_{n-1}$ in $\text{(2a)}$
$\text{(3b):}$ substitute $n\mapsto n-1$ in $\text{(2a)}$ to get $8b_{n-2}=a_{n-1}-3a_{n-2}$
$\phantom{\text{(3b):}}$ then mulitiply by $3$ to get $24b_{n-2}=3a_{n-1}-9a_{n-2}$
$\phantom{\text{(3b):}}$ then use for $24b_{n-2}$ in $\text{(3a)}$

Since $\text{(2a)}$ says $b_n=\frac{a_{n+1}-3a_n}8$, we see that $b_n$ also satisfies the recurrence in $(3)$: $b_n=6b_{n-1}-b_{n-2}$.

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An Observation

Since $a_0=1$ and $a_1=3$ and $a_n$ satisfies the recursion in $(3)$, we get the first several terms $$ 1,3,17,99,577,3363,\dots\tag4 $$ a bit of observation shows that this sequence is $$ 0^2+1,2^2-1,4^2+1,10^2-1,24^2+1,58^2-1,\dots\tag5 $$ The terms which are squared appear to follow the recursion $c_n=2c_{n-1}+c_{n-2}$.

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Proving the Observation

Suppose that $c_n=2c_{n-1}+c_{n-2}$. Then $$ \begin{align} c_n^2 &=(2c_{n-1}+c_{n-2})^2\tag{6a}\\ &=4c_{n-1}^2+c_{n-2}^2+4c_{n-1}c_{n-2}\tag{6b} \end{align} $$ Furthermore, $$ \begin{align} c_nc_{n-1} &=(2c_{n-1}+c_{n-2})c_{n-1}\tag{7a}\\ &=2c_{n-1}^2+c_{n-1}c_{n-2}\tag{7b} \end{align} $$ Therefore, $$ \begin{align} c_n^2-4c_{n-1}^2-c_{n-2}^2&=4c_{n-1}c_{n-2}\tag{8a}\\ c_{n-1}^2-4c_{n-2}^2-c_{n-3}^2&=4c_{n-2}c_{n-3}\tag{8b}\\ c_n^2-5c_{n-1}^2+3c_{n-2}^2+c_{n-3}^2&=8c_{n-2}^2\tag{8c}\\ c_n^2-5c_{n-1}^2-5c_{n-2}^2+c_{n-3}^2&=0\tag{8d} \end{align} $$ Explanation:
$\text{(8a):}$ apply $(6)$
$\text{(8b):}$ apply $(6)$
$\text{(8c):}$ subtract $\text{(8b)}$ from $\text{(8a)}$ and apply $(7)$
$\text{(8d):}$ cancel

Thus, $c_n^2$ satisfies the linear recurrence in $\text{(8d)}$. The characteristic polynomial of the linear recurrence in $\text{(8d)}$ can be factored as $$ x^3-5x^2-5x+1=\underbrace{\ (x+1)\ \vphantom{\left(x^2\right)}}_{(-1)^n}\underbrace{\left(x^2-6x+1\right)}_{a_n,b_n}\tag9 $$ That is, if $c_n=2c_{n-1}+c_{n-2}$, then $c_n^2$ is a linear combination of $a_n$, $b_n$, and $(-1)^n$.

Suppose that $c_0=0$, $c_1=2$, and $c_2=4$. Then, solving for the coefficients of $a_n$, $b_n$, and $(-1)^n$, we get $$ c_n^2=a_n-(-1)^n\tag{10} $$ Thus, we have shown that our observation in $(5)$ is true: $$ a_n=c_n^2+(-1)^n\tag{11} $$ where $c_0=0$, $c_1=2$, and $c_n=2c_{n-1}+c_{n-2}$.

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Answering the Question

Looking at the odd indexed terms, we get $$ \begin{align} a_{2n+1} &=c_{2n+1}^2-1\tag{12a}\\ &=(c_{2n+1}-1)(c_{2n+1}+1)\tag{12b} \end{align} $$ as proposed in the question.