2
$\begingroup$

$(a)$ Consider the recurrence relation $a_{n+2}a_n = a^2 _{n+1} + 2$ with $a_1 = a_2 = 1$.

$(i)$ Assume that all $a_n$ are integers. Prove that they are all odd and the integers $a_n$ and $a_{n+1}$ are coprime for $n \in \mathbb N$

$(ii)$ Assume that the set $\{a_n , a_{n+1} , a_{n+2}\}$ is pairwise coprime for $n \in \mathbb N$. Prove that all $a_n$ are integers by induction.

$(b)$ Consider the recurrence relation $a_{n+2}a_n = a^2_{n+1} + 1$ with $a_1 = 1, a_2 = 2$ and compare this sequence to the Fibonacci numbers. What do you find? Formulate it as a mathematical statement and prove it.

  • 1
    (_ii_) is kind of strange : the notion of coprimality doesn't really make sense for non-integers, so in some sense it's making the statement: 'assume $\{a_n, a_{n+1}, a_{n+2}\}$ are integers for all $n\in\mathbb{N}$; prove that all $a_n$ are integers.'2012-11-13
  • 0
    I noticed that in part (b) the $a_6 = 121/2$ is not an integer.2012-11-13
  • 0
    @spernerslemma It's subtle but the recurrence is changed in part b. (+2 becomes +1)2012-11-13
  • 0
    @EuYu, so it's going odd, even, odd, even, .. and so we get odd/even! thanks.2012-11-13
  • 0
    @spernerslemma Yea. From a brief inspection, it seems like the sequence of odd Fibonacci numbers $F_{2n+1}$2012-11-13

1 Answers 1