it is well known: if $F_{1}=1,F_{2}=1$,and $F_{n+1}=F_{n}+F_{n-1}$.use $$F_{m+n}=F_{m}F_{n+1}+F_{m-1}F_{n}$$ I can prove $(F_{m},F_{n})=F_{(m,n)}$
also I have solve this following problem $a_{1}=a_{2}=1$,and $p,q$ be postive integer,and such $a_{n+2}=pa_{n+1}+qa_{n}$ if $$p=1\Longleftrightarrow (a_{m},a_{n})=a_{(m,n)}$$ proof:since $$a_{3}=p+q,a_{4}=p^2+pq+q,a_{5}=p^3+p^2q+2pq+q^2,a_{6}=p^4+p^3q+3p^2q+2pq^2+q^2$$ if $(a_{m},a_{n})=a_{(m,n)}$ so $$1=a_{(3,4)}=(a_{3},a_{4})=(p+q,p^2+pq+q)\Longrightarrow (p,q)=1$$ and note $a_{3}=(a_{3},a_{6})\Longrightarrow a_{3}|a_{6}$，so $$(p+q)|(p^2q+q^2)=q(p+q)+pq(p-1)$$ so $p=1$
and other hand we if $p=1$,use 1 we have $$a_{m+n}=a_{m+1}a_{n}+qa_{m}a_{n-1}$$ and it is easy to show $(a_{n+2},a_{n+1})=(qa_{n},a_{n+1})=(a_{n},a_{n+1})$ because $(a_{n+2},q)=(a_{n+1},q)=\cdots=(1,q)=1$ so we have$(a_{n+1},a_{n})=1$, then $$(a_{m+n},a_{n})=(a_{m}a_{n+1}+qa_{m-1}a_{n},a_{n})=(a_{m},a_{n})=\cdots=(a_{(m,n)},a_{(m,n)})=a_{(m,n)}$$ then we have $$(a_{m},a_{n})=a_{(m,n)}$$
My question: let $a,b,x,y$ be postive integers,and sequence $\{a_{n}\}$ such $a_{1}=a,a_{2}=b$,and $$a_{n+1}=xa_{n}+ya_{n-1}$$ find the $a,b,x,y$ such $(a_{m},a_{n})=a_{(m,n)}$
