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