The trick is to use the bound $\gcd(x,y)\leq |x-y|$.

we can observe that $a_{i+1}-a_i=gcd(a_{i-1},a_{i-2}),~~\forall i\in\mathbb{N}-\{1,2\}~~~(1)$ $\Rightarrow a_{i+1}-a_i=gcd(a_{i-2},a_{i-2}+gcd(a_{i-3},a_{i-4})=gcd(a_{i-2},gcd(a_{i-3},a_{i-4}))$ thus $a_{i+1}-a_i\leq gcd(a_{i-3},a_{i-4})=a_{i-1}-a_{i-2},~~\forall i\in\mathbb{N}-\{1,2,3,4\}~~~(*)$ let$x_i=a_{2i}-a_{2i-1}~~\forall i\in\mathbb{N}$ from $(*)$ we'll get that $x_n$ is a non increasing sequence since we can obviously see that $x_n$ is a sequence of positive integer there exist $r\in\mathbb{N}$ such that $x_{r+i}=p,~\forall i\in\mathbb{N}\cup{0}$ where $p\in\mathbb{N}$ in a similar way let $y_i=a_{2i+1}-a_{2i}~~\forall i\in\mathbb{N}$ we'll get that there exist $s\in\mathbb{N}$ such that $y_{s+i}=q,~\forall i\in\mathbb{N}\cup{0}$ where $q\in\mathbb{N}$ let $z=max\{r,s\}+3$ thus $y_{z+i}=q,x_{z+i}=p,~~\forall i\in\mathbb{N}\cup{0}$ thus $a_{2z+1}-a_{2z}=q$ and $a_{2z+2}-a_{2z+1}=p$ $\Rightarrow a_{2z+2}=a_{2z}+p+q,a_{2z+1}=a_{2z}+q$ from $(1)$ we'll get that $a_{2z+2}-a_{2z+1}=gcd(a_{2z},a_{2z-1})=p\Rightarrow p|a_{2z}$ hence ,$ x_{z+2}=a_{2z+4}-a_{2z+3}=gcd(a_{2z+2},a_{2z+1})=gcd(a_{2z}+p+q,a_{2z+1})=p$ thus $p|a_{2z}+p+q\Rightarrow p|q~~~(\because p|a_{2z})$ in a similar way we'll get that $y_{z+1}=a_{2z+3}-a_{2z+2}=gcd(a_{2z+1},a_{2z})=q\Rightarrow q|a_{2z}$ consider $y_{z+2}=a_{2z+5}-a_{2z+4}=gcd(a_{2z+3},a_{2z+2})=q\Rightarrow q|a_{2z+2}$ which is $q|a_{2z}+p+q\Rightarrow q|p~~~(\because q|a_{2z})$ from $p|q,~q|p$ and $p,q\in\mathbb{N}$ we can conclude that $p=q$ therefore $x_{z+i}=y_{z+i},~~\forall i\in\mathbb{N}\cup\{0\}$ thus $a_{i+1}-a_{i}$ is constant for all posotive integer $i$ more than $2z-2$ $\therefore$ there exist two positive integer numbers $N, M$, such that $a_{n+1}-a_n=M$ for all $n \geq N~~\square$ lmk if there is any mistakes