Solve the equation $5x^2+5y^2-8xy-2x-4y+5=0$ While I was solving the question given by Harsh Kumar I could not find the real roots just because I changed the sign of $5y^2$. I also proved that there don't exist any real solution of that equation.

The equation I made by mistake was: $$5x^2+5y^2-8xy-2x-4y+5=0.$$

Please help me to find the solutions if exist.

We may rewrite this as
$$
4x^2 - 8xy + 4y^2 + x^2 -2x + 1 + y^2 - 4y + 4 = 0\\
(2x - 2y)^2 + (x-1)^2 + (y-2)^2 = 0
$$
and we see that this is a sum of three (real) squares that equals $0$. If the sum of three squares equals $0$, then that means that each of the squares are zero. But $(x-1)^2 = 0$ means $x = 1$ and $(y-2)^2 = 0$ means $y = 2$, and then $(2x-2y)^2 \neq 0$. Therefore there can be no real solutions to this polynomial equation.
