Find all postive integer numbers $x,y$,such $x+y+1$ divides $2xy$ and $x+y-1$ divides $x^2+y^2-1$ Find all postive integer $x$ and $y$ such that
$x+y+1$ divides $2xy$ and $x+y-1$ divides $x^2+y^2-1$
My try: since
$$(x+y)^2-2xy=x^2+y^2$$
I know this well know reslut:
$$xy|(x^2+y^2+1),\Longrightarrow \dfrac{x^2+y^2+1}{xy}=3,x,y\in N$$
and such condition $$(x,y)=(u_{n},u_{n+1})$$
where
$$u_{0}=u_{1}=1,u_{n+2}=3u_{n+1}-u_{n}$$
so I can't,Thank you 

Note that since $(x+y+1)(x+y-1)=(x^2+y^2-1)+2xy$,
$$
x+y-1\mid x^2+y^2-1\iff x+y-1\mid2xy\tag{1}
$$
Since
$$
x+y+1\mid2xy\tag{2}
$$
and $(x+y-1,x+y+1)\mid2$, $(1)$ and $(2)$ imply
$$
(x+y)^2-1\mid4xy\tag{3}
$$
However, because $4xy\le(x+y)^2$, $(3)$ implies $(x+y)^2-1=4xy$, that is,
$$
(x-y)^2=1\tag{4}
$$
Wlog, assume $y=x+1$, then
$$
x+y-1=2x\mid2x(x+1)=2xy\tag{5}
$$
and
$$
x+y+1=2(x+1)\mid2x(x+1)=2xy\tag{6}
$$
Thus, the solutions are $\{(x,y):(x-y)^2=1\}$
