if $a_{n+1}=xa_{n}+ya_{n-1}$ then have this results :$a_{m+n}=Xa_{m}a_{n+1}+Ya_{m-1}a_{n}$？ it is well  known:
if $F_{1}=1,F_{2}=1$,and $F_{n+1}=F_{n}+F_{n-1}$.then we have
$$F_{m+n}=F_{m}F_{n+1}+F_{m-1}F_{n}$$
prove 1
and I have know this :if $a_{0}=0,a_{1}=1$ and $x,y$ be give postive integers,and $$a_{n+1}=xa_{n}+ya_{n-1}$$
then we have
$$a_{m+n}=a_{m}a_{n+1}+ya_{m-1}a_{n}$$
prove 2
My question is
if the sequence such $a,b,x,y$ be give postive integers,and such that  $a_{1}=a,a_{2}=b,a,b\in N^{+}$,and $$a_{n+1}=xa_{n}+ya_{n-1}$$,then is also have this  Similar results
$$a_{m+n}=Xa_{m}a_{n+1}+Ya_{m-1}a_{n}?$$
if there exist,Find the $X,Y$?Thanks
 A: So, the answer is indeed, yes, such a relation exists (with some caveats), and we can prove it via induction, just as in the 1st answer you have linked in the question.
MAJOR EDITS: My original answer was incorrect, as pointed out by @mathlove. Here is a revised answer.
First, let us prove the induction step. For it, assuming the relation is true upto $(m,n)$, we have to prove it for $(m+1,n)$ and $(m,n+1)$, just as mentioned in the $1^{st}$ answer.
For $(m+1,n)$, we have
\begin{align}
a_{m+1+n} &= xa_{m+n}+ya_{(m-1)+n} \\
&= x(Xa_ma_{n+1}+Ya_{m-1}a_n) + y(Xa_{m-1}a_{n+1}+Ya_{m-2}a_n) \\
&= Xa_{n+1}(xa_m+ya_{m-1}) + Ya_n(xa_{m-1}+ya_{m-2}) \\
&= Xa_{m+1}a_{n+1} + Ya_ma_n
\end{align}
Similarly we can do so for $(m,n+1)$. Thus, the inductive step is proved.
For the base case, as the induction relation involves $m-1$ and $n$, the lowest values that these can have is $1$ (As $a_0$ is not defined). So, the base case for the induction must be for $(m,n) = (2,1)$.
For the base case to hold, the following relation must hold -
\begin{align}
a_{2+1} &= Xa_2a_{1+1} + Ya_{2-1}a_1 \\
\implies a_3 &= Xb^2 + Ya^2
\end{align}
From the given recursion relation, we get
\begin{align}
a_3 &= xa_2 + ya_1 \\
&= xb + ya
\end{align}
Hence, the given relation holds iff $X,Y$ satisfy
$$Xb^2 + Ya^2 = xb + ya$$
Addendum: I added this as this was part of my original answer, and I wanted to show where I originally went wrong.
Now, let us determine what restrictions $X$ and $Y$ must have, if such a result was to exist.
Let $m=2$. Then, the induction gives
\begin{align}
a_{n+2} &= Xa_2a_{n+1}+Ya_1a_n \\
&= (Xb) a_{n+1} + (Ya) a_n
\end{align}
Now, we have already been given that $a_{n+2}=xa_{n+1}+ya_{n}$. Thus, we get
\begin{align}
(Xb) a_{n+1} + (Ya) a_n &= xa_{n+1}+ya_{n}\\
\implies (Xb-x)a_{n+1} &= (y-Ya)a_n
\end{align}
Now, if either $X = \frac{x}{b}$ or $Y = \frac{y}{a}$, then this relation shows that the other must be zero as well. Hence, we have either $(X,Y) = \left(\frac{x}{b}, \frac{y}{a}\right)$, or
\begin{align}
a_{n+1} &= \frac{y-Ya}{Xb-x}a_n \\
\implies a_n &= \left(\frac{y-Ya}{Xb-x}\right)^{n-1} a_1
\end{align}
Note: We can easily see that in both the cases, $Xb^2 + Ya^2 = xb + ya$ holds (as it should).
