[nooonuii]

ข้อความเดิมเขียนโดยคุณ dektep

4.The sequence $(x_n)$ is defined as; $x_1 = a$, $x_2 = b$ and for all positive integer $n$, $x_{n + 2} = 2008x_{n + 1} - x_n$. Prove that there are some positive integers $a,b$ such that $1 + 2006x_{n + 1}x_n$ is a perfect square for all positive integer $n$.

Hint:

มองหา invariant ให้เจอ แล้วจะรู้ว่า $a,b$ ควรเป็นอะไร

My Solution :

ให้ $a=1,b=2008$ จะำพิสูจน์ว่า $$1+2006x_nx_{n+1}=(x_{n+1}-x_n)^2$$
Proof :
$(x_{n+1}-x_n)^2-2006x_nx_{n+1}-1=x_{n+1}^2-2008x_nx_{n+1}+x_n^2-1$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=(x_{n+1})(x_{n+1}-2008x_n)+x_n^2-1$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=(2008x_n-x_{n-1})(-x_{n-1})+x_n^2-1$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=x_n^2-2008x_{n-1}x_n+x_{n-1}^2-1$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\vdots$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=x_2^2-2008x_1x_2+x_1^2-1$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=0$