Show that the sequence $a_n$ is eventually periodic 
Suppose that $a_0, a_1, \cdots $ and $b_0, b_1, \cdots$ are two sequences of positive integers such that $a_0, b_0 \ge 2$ and $$ a_{n+1} = \gcd{(a_n, b_n)} + 1, \qquad b_{n+1} = \operatorname{lcm}{(a_n, b_n)} - 1. $$ Show that the sequence $a_n$ is eventually periodic; in other words, there exist integers $N \ge 0$ and $t > 0$ such that $a_{n+t} = a_n$ for all $n \ge N$.

Hint: Try to prove the sequence is bounded above.
Can someone explain to me how showing the sequence bounded above completed the proof? I have not got that part.
Here is my progress.
Note that $a_i\ge 2$. Moreover, $a_i\ge a_{i+1}$ whenever $a_i\not \mid b_i$. When $a_i|b_i\implies a_{i+1}=a_i+1, b_{i+1}=b_i-1$.
Call a number $a_i$ good if $a_i<a_{i+1}$. Note that a sequence can not have an infinite subsequence containing good numbers.
Hence, there will be a subsequence such that $$(a,b)\rightarrow (a-c,d+c)\rightarrow \dots \rightarrow (a-1,d+1) \rightarrow (a,d)\rightarrow (X,Y).$$
(else the sequence $a_i$ would be bounded above by $a$, which would solve the problems)
So, now, consider the sequence $$(a,b)\rightarrow (a-c,d+c)\rightarrow \dots \rightarrow (a-1,d+1) \rightarrow (a,d)\rightarrow (X,Y).$$
Note that $$\gcd(a,b)+1=a-c, \text{lcm}(a,b)-1=d+c\implies $$
Note that $$(a,d)=(a+d,a), a-c|d+c\implies a-c|a+d.$$
I want to try and prove that $X=a-c$ i.e show that $(a,d)=a-c+1$. But I do not know how to. Any hints or solution, in this direction?
 A: I am going just to review and post some mentioned ideas together, with more explanation, in a way that might be helpful to readers:
What we need is to show that the sequence $a_0, a_1, a_2, ...$ is bounded. Let's call $a_i$ "a peak" if $a_i\geq a_{i+1}$. As it is noticed above, if $a_i$ is not a peak, then $(a_{i+1}, b_{i+1})=(a_i+1,b_i-1)$.
Now, assume $a_m$ and $a_n$ are two consecutive peaks ($n\gt m $).
First we prove that $a_m\ge a_n$.
If $a_m\lt a_n$, then $(a_{m+1},b_{m+1})=(d+1, dxy-1)$ where $a_m=dx, b_m=dy $, and $ gcd(x,y)=1$ (simply because $a_{m+1}$ is not a peak).  Since $a_m\lt a_n$ (by assumption), we can arrive at an index, namely $j$, where $m\lt j\lt n$, and $(a_j,b_j)=(dx, dxy-dx+d).$ The reason behind this claim is obvious. Just notice that none of the indices between $m$ and $n$ is a peak, so at each step $a_i$ is increased by $1$ and $b_i$ is decreased by $1$ and this process can be repeated $dx-d-1$ times as $dx= a_m\lt a_n$. But this is a contradiction because, in this manner,  $a_{j+1}=d+1$, which means $a_j$ is a peak (We had supposed $a_m$ and $a_n$ were two consecutive peaks).
Therefore, the subsequesnce of all peaks will be eventually constant, and this imply that the sequence $a_0, a_1, a_2, ...$ is bounded because it is impossible to have infinite non-peaks between two consecutive peaks (otherwise for some $i$, $b_i\lt2$).
Now, let $C$ be the maximum of the sequence $a_0, a_1, a_2, ...$ and $N=C!$, then the pair $(a_n, b_n\; mod\;N)$ determines the pair $(a_{n+1}, b_{n+1}\; mod\;N)$ (Just because $gcd(a_n,b_n)=gcd(a_n, b_n\; mod\; N)$). As the sequence  $a_0, a_1, a_2, ...$ is bounded and $b_n\; mod\;N$ is limited, we will finally get the pair $(a_n, b_n\; mod\;N)$ somewhere repeated, which means the sequence is periodic.
