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 
 A: 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 $x=y-1$
A: 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 $\{(x,y):(x-y)^2=1\}$
A: $$(x+y+1)(x+y-1)-1(x^2+y^2-1)=2xy$$
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 $\{(x,y)\in\Bbb N^2\mid |x-y|=1\}$
