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 

Noting that 
$\left(x+y+1\right)\left(x+y-1\right)=x^2+y^2-1+2xy\quad * $ 
Use the above expression to prove that $x+y-1$ divides $2xy$ and $x+y+1$ divides $x^2+y^2-1$
Note that $x+y-1$ and $x+y+1$ have the same parity.
Let both of them be odd, then they are coprime
$\implies \left(x+y+1\right)\left(x+y-1\right)$  divides each of $2xy$ and $x^2+y^2-1$
Let the quotient of the division be $a$ and $b$ respectively. Use $*$ to arrive at $a+b=1$

 $$a\left(x+y+1\right)\left(x+y-1\right)=2xy$$ and $$b\left(x+y+1\right)\left(x+y-1\right)=x^2+y^2-1$$ adding both of them gives $\left(a+b\right)\left(x+y+1\right)\left(x+y-1\right)=x^2+y^2-1+2xy=\left(x+y+1\right)\left(x+y-1\right)$ as we are speaking of $2xy$... being divided by $x+y+1$ or $x+y-1$ and as we don't say is divisible by zero, we can neglect the case in which $\left(x+y+1\right)\left(x+y-1\right)=0$. Which yields the condition $a+b=1$

let both of them be even.
$\implies \frac{\left(x+y+1\right)\left(x+y-1\right)}{2}$ divides each of $2xy$ and $x^2+y^2-1$ . Let the quotient of the division be $c$ and $d$ respectively. Use $*$ to arrive at $c+d=2$.

 $$c\frac{\left(x+y+1\right)\left(x+y-1\right)}{2}=2xy$$ and $$d\frac{\left(x+y+1\right)\left(x+y-1\right)}{2}=x^2+y^2-1$$ adding both of them gives $\left(a+b\right)\frac{\left(x+y+1\right)\left(x+y-1\right)}{2}=x^2+y^2-1+2xy=\left(x+y+1\right)\left(x+y-1\right)$ as we are speaking of $2xy$... being divided by $x+y+1$ or $x+y-1$ and as we don't say is divisible by zero, we can neglect the case in which $\left(x+y+1\right)\left(x+y-1\right)=0$. Which yields the condition $a+b=2$

Since $a,b,c$ and $d$ are positive integers, there cannot exist a case in which $a+b=1$. And $c=d=1$. This causes either of $x+y-1$ or $x+y+1$ zero which is absurd as we are talking about division.
For the second case we have $x^2+y^2-1=2xy=\frac{\left(x+y+1\right)\left(x+y-1\right)}{2}$
Which gives $\left(x-y\right)^2=1$ $\implies x=y+1$ or $x=y-1$
