Challenge problem #1 Solve 2x^2y^2−2xy+x^2+y^2−2x−2y+3=0

[Olinguito]

Hi all.

I would like to post some challenge problems from time to time. I’ll start with a simple one. :)

Find all real numbers $x,y$ satisfying the following equation:
$$2x^2y^2-2xy+x^2+y^2-2x-2y+3=0.$$

 

[Opalg]

Hi all.

I would like to post some challenge problems from time to time. I’ll start with a simple one. :)

Find all real numbers $x,y$ satisfying the following equation:
$$2x^2y^2-2xy+x^2+y^2-2x-2y+3=0.$$

Hi Olinguito, and welcome to MHB! We look forward seeing your problems.
[sp]$2x^2y^2-2xy+x^2+y^2-2x-2y+3 = 2(xy-1)^2 + (x+y-1)^2$. If that is zero then $xy=1$ and $x+y=1$. So $x$ and $y$ are the roots of $\lambda^2 - \lambda + 1 = 0$. But that equation has no real roots, so the given equation has no real solutions.[/sp]

 

[Olinguito]

Thanks Opalg – and great work! :D

I should have a second problem ready soon.

 

[I like Serena]

Hi all.

I would like to post some challenge problems from time to time. I’ll start with a simple one. :)

Find all real numbers $x,y$ satisfying the following equation:
$$2x^2y^2-2xy+x^2+y^2-2x-2y+3=0.$$

Welcome Olinguito!

Spoiler

$$2x^2y^2-2xy+x^2+y^2-2x-2y+3 = (xy)^2 + (xy-1)^2+(x-1)^2+(y-1)^2=0$$
A sum of squares is 0 if and only if all individual squares are 0.
So $x=1,y=1$,and $xy=0$, which is a contradiction.
Therefore there are no solutions.

 

[Olinguito]

Thanks, I like Serena! Great work as well. :D