<s>[INST] 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 
 [/INST] $$(x+y+1)(x+y-1)-1(x^2+y^2-1)=2xy$$
If we assume $x\equiv y\pmod 2$, then $\text{lcm}(x+y+1,x+y-1)=(x+y)^2-1$ since the expressions are coprime.  If $x+y-1\mid x^2+y^2-1$, the above equality would imply that $x+y-1\mid 2xy$, but since $2xy<(x+y)^2-1$ ($x,y>0$), this implies $x+y+1\not \mid 2xy$.
If $x\not\equiv y\pmod2$, then $\text{lcm}(x+y+1,x+y-1)=\frac{(x+y)^2-1}2$.  In order to construct a similar argument to before, we need $$\begin{align} \frac{(x+y)^2-1}{2}&>2xy\\
(x+y)^2-1&>4xy\\
(x+y)^2-4xy&>1\\
(x-y)^2&>1\\
|x-y|&>1 \end{align}$$
This leaves us with two cases: $x=y$ and $x=y+1$.  We can dismiss the first since it implies $x\equiv y\pmod 2$.  So we want $$(y+1)+y-1=2y\mid (y+1)^2+y^2-1=2y^2+2y$$ Which is true.
We also need $$(y+1)+y+1=2y+2\mid 2(y+1)y$$ Which is true for all $y$.

In conclusion, your solution set for the problem is $\{(x,y)\in\Bbb N^2\mid |x-y|=1\}$
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