# Solve the following Diophantine equation.

Find all integral pairs (x,y) such that - $$( xy - 1)^2 = (x +1)^2 + ( y+1)^2$$

My Approach :

I just expanded this equation and wrote it in another form - $$\frac{(xy+1)(xy-1)}{(x+y)}-2=x+y$$ and from this we can say that $(x+y)|(xy+1) \ \mathrm{or}\ (x+y)|(xy-1)$ . But i don't know how to solve it further. Please help me with this.

Hint:

$$x^2(y^2-1)-2x(1+y)-(y+1)^2=0$$

What if $y+1=0?$

Else $$x^2(y-1)-2x-(y+1)=0$$

Method$\#1:$

$$x=\dfrac{2\pm\sqrt{4+4(y^2-1)}}{2(y-1)}=\dfrac{1\pm y}{1+y}$$

Now $\dfrac{1-y}{1+y}=\dfrac2{1+y}-1$

$\implies1+y$ must divide $2$

Method $\#2:$

By symmetry, $x+1$ must be a factor of $$x^2(y-1)-2x-(y+1)$$

What is the quotient?

Or $$x^2y^2 = (x+y)^2+2(x+y)+1 \implies x^2y^2 = (x+y+1)^2$$

1. case: $$xy =x+y+1\implies (x-1)(y-1)=2\implies ....$$

2. case: $$xy =-x-y-1\implies (x+1)(y+1)=0\implies ....$$