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

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The equation $2x^2y^2-2xy+x^2+y^2-2x-2y+3=0$ has been analyzed and determined to have no real solutions. The equation can be rewritten as $2(xy-1)^2 + (x+y-1)^2$. For this expression to equal zero, both conditions $xy=1$ and $x+y=1$ must be satisfied. However, the resulting quadratic equation $\lambda^2 - \lambda + 1 = 0$ has no real roots, confirming the absence of real solutions.

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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:
$$2x^2y^2-2xy+x^2+y^2-2x-2y+3=0.$$
 
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Olinguito said:
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]
 
Thanks Opalg – and great work! :D

I should have a second problem ready soon. :cool:
 
Olinguito said:
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!
$$2x^2y^2-2xy+x^2+y^2-2x-2y+3 = (xy)^2 + (xy-1)^2+(x-1)^2+(y-1)^2=0$$
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.
 
Thanks, I like Serena! Great work as well. :D
 

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