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.




What if $y+1=0?$

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


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.