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Find all real pairs $(x,y)$ that satisfy the equation $$5x^2 + 5y^2 + 8xy + 2y - 2x + 2 = 0$$

My attempt:

I thought that it could be solved by factoring everything to construct 2 different equations $(e_1)(e_2) = 0$, however, it seems that the factoring doesn't work too well:

The factored equation should be of form: $(ax+by+c)(lx+my+n)=0$, so it is possible to create a system of equations from that:

$Al = 5$

$bm = 5$

$Am + bl = 8$

$An + cl = -2$

$bn + cm = 2$

$cn = 2$

However, after a few hours of meddling with it, I had found no solutions. That is when I started to wonder that perhaps I am doing something wrongly.

P.S. The problem is supposed to be solvable by a high school student, not a graduate in mathematics.

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Note that:-

$$5x^2 + 5y^2 + 8xy + 2y - 2x + 2 = 0$$

can be written as-

$$(2x+2y)^2+(x-1)^2+(y+1)^2=0\space\space\space\space\space\space\space\space\text{[How?]}$$

Now,use the fact that square of any number is non-negative.So,the above equation is only possible when each squared term is equal to $0$.Now,solve!!

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solving this equation for $y$ and we get $$y=\frac{1}{5}\left(-1-4x\pm\sqrt{-1+2x-x^2}\right)$$ the radicand is $$-(1-2x+x^2)=-(x-1)^2$$ thus we get $$x=1$$ and $$y=-1$$

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