1.(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 $\left\{ a_n,\ a_{n+1},\ a_{n+2}\right\}$ 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.
I have no idea where to start with 1(ai) but with 1(aii) i have started with:
Given that $\left\{a_n,\ a_{n+1},\ a_{n+2}\right\}$ is pairwise coprime you get $gcd(a_n,a_{n+1})=1,\ gcd(a_n,a_{n+2})=1,\ gcd(a_{n+1},a_{n+2})=1$
Using the initial terms, you can do base induction on $a_{n+2} = \frac{a^2_{n+1}+2}{a_n}$ to prove whether the next terms will be integers.
$a_3 = \frac{1^2+2}{1}$ $a_4 = \frac{3^2+2}{1}$ $a_5 = \frac{11^2+2}{3}$
gives:
$a_1=1,\ a_2 = 1,\ a_3 = 3,\ a_4 = 11,\ a_5 = 41$
which are all integers. So the base induction is correct.
Now for the inductive step $n=k, k \rightarrow k+1$
$a_{k+3} = \frac{a^2_{k+2}+2}{a_{k+1}}$. I am not sure how to prove this for the inductive step.
Also for (b) how do you formulate a mathematical statement and prove it?