0

Let's be sequence $ (u_n) $ is defined by $u_1=1$, $u_{n+1} =2 u_n+\sqrt{3 u_n^2+6}, \quad \forall n \geq 1.$ Prove that all its members are positive integers.

I tried by using matrix. Put $ v_n = \sqrt{3 u_n^2+6} $, then $ v_n^2 - 3u_n^2 = 6 $ and $$ u_{n+1}=2u_n + v_n, \quad \forall n \geq 1. $$ We have \begin{align*} v_{n+1}^2 &= 3u_{n+1}^2 + 6\\ &= 3(2u_n + v_n)^2 +v_n^2 - 3u_n^2\\ &= 9u_n^2 + 12 u_n v_n + 4v_n^2\\ &= (3u_n + 2 v_n)^2\\ \end{align*} Then, $ v_{n+1} = 3u_n + 2 v_n. $ We have $$ \left( \begin{array}{c} u_1 \\ v_1 \\ \end{array}\right) = \left( \begin{array}{c} 1 \\ 3 \\ \end{array}\right), \quad \left( \begin{array}{c} u_{n+1} \\ v_{n+1} \\ \end{array}\right) = \left( \begin{array}{cc} 2& 1 \\ 3 &2 \\ \end{array}\right)\left( \begin{array}{c} u_n \\ v_n \\ \end{array}\right)$$

Thus all the numbers $u_ n$ and $v_ n , n ≥ 2,$ are positive integers.

How can I solve this prolbem without using matrix?

minhthien_2016
  • 724
  • 1
    The recursion in that Q&A is slightly different, but the identical techniques work in your case and show that $u_{n+1} = 4 u_n - u_{n-1}$. – Martin R Nov 13 '19 at 08:57
  • In addition, you don't have to write it using a matrix. You already have the recurrence relation (over 2 variables), so checking that the base cases are integers is sufficient. – Calvin Lin Nov 13 '19 at 09:17

0 Answers0