2002
Problem - 3173
Let $\alpha$ and $\beta$ be the two roots of the equation $x^2 -x - 1=0$. If $$a_n = \frac{\alpha^n - \beta^n}{\alpha -\beta}\quad(n=1, 2, \cdots)$$ Show that
- For any positive integer $n$, it always hold $a_{n+2}=a_{n+1}+a_n$
- Find all positive integers $a, b$ $( a < b )$ satisfying $b\mid a_n-2na^n$ holds for any positive integer $n$
i) Because $\alpha$ and $\beta$ are the two roots of $x^2 -x-1=0$, we have
\begin{align}
\alpha^2 = \alpha +1 &\implies \alpha^{n+2}=\alpha^{n+1}+\alpha^n\\
\beta^2 = \beta +1 &\implies \beta^{n+2}=\beta^{n+1}+\beta^n
\end{align}
\begin{align}
\therefore\quad a_{n+2}= &\frac{\alpha^{n+2}-\beta^{n+2}}{\alpha - \beta}\\
=&\frac{(\alpha^{n+1}+\alpha^n)-(\beta^{n+1}+\beta^n)}{\alpha-\beta}\\
=&\frac{(\alpha^{n+1}-\beta^{n+1})+(\alpha^{n}-\beta^n)}{\alpha-\beta}\\
=&\frac{(\alpha^{n+1}-\beta^{n+1})}{\alpha-\beta}+\frac{(\alpha^{n}-\beta^n)}{\alpha-\beta}\\
=&a_{n+1} + a_n
\end{align}
ii) Clearly, we have $a_1=1$, $a_2=\alpha+\beta=1$, and therefore, $a_3=a_1+a_2=2$.
From the conditions, we have $b\mid (a_1 - 2a) = (1-2a)$. Meanwhile $$0 < a < b \implies 1 \le 2a-1 < 2b-1 < 2b$$
Therefore, it must hold that $$b=2a-1$$
Setting $n=3$ in the given condition leads to $$b\mid a_3 -6a^3= 2 -6a^3\implies b\mid 6a^3-2$$
Because $b=2a-1$, therefore
\begin{align}
&2a-1 \mid 6a^3-2=3a^2(2a-1)+(3a^2-2)\\
\implies & 2a-1\mid 3a^2-2\\
\implies & 2a-1\mid 6a^2-4 = (3a+1)(2a-1)+(a-3)\\
\implies & 2a-1 \mid a-3\\
\implies & 2a-1 \mid 2a-6 =(2a-1)-5\\
\implies & 2a-1 \mid 5\\
\implies & a = 1, 3
\end{align}
When $a=1$, $b=2a-1=1$ violates the condition $a < b$. When $a=3$, $b=5$. Hence, the answer is $$(a, b)=\boxed{(3, 5}$$