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I have no idea how to solve the system below

\begin{cases} f_x = - 2 x y + y^{2} + 1 = 0 \\ f_y = - x^{2} + 2 x y - 1 = 0 \end{cases}

I began by using this linear combination $c_1 f_x + c_2 f_y = 0$ for $c_1,c_2 \in \mathbb{R}$. I set $c_1 = 1$ and $c_2 = 1$ and I got $$y^2 - x^2 - 0$$. It seems to be that this equation have infinite solutions: $(-1,-1), (0,0), (1,1), (2,2), ...$. Is my reasoning correct? When I feed the system to computer, I got $\left(1, 1\right)$ and $\left(-1, -1\right)$ as only solutions. I don't understand why.

Ray Siplao
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    While any solution to the original equations must also be a solution to $y^2-x^2=0$, not every solution to the latter is also a solution to the original system. For example, as you note $x=y=0$ is a solution to the latter; but it is not a solution to $f_x=0$. You have more work still to be done. – Arturo Magidin Oct 04 '21 at 04:32

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$$f_x = - 2 x y + y^{2} + 1 = 0 \tag 1$$ $$f_y = - x^{2} + 2 x y - 1 = 0 \tag 2$$

Since $y=0$ cannot be a solution (check it for $(1)$), then $$f_x =0 \implies x=\frac{y^2+1}{2 y}\tag 3$$ Plug in $(2)$ to get $$y^2-\frac{\left(y^2+1\right)^2}{4 y^2}=0\implies 3 y^4-2 y^2-1=0 \tag 4$$ $$3 y^4-2 y^2-1=(y-1) (y+1) \left(3 y^2+1\right)$$ So, the only solutions of $(4)$ are $y=\pm1$ and, back to $(3)$, $x=\pm 1$.

Claude Leibovici
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