# How to solve the below system of equations?

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.

• 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. Oct 4 '21 at 4:32

$$f_x = - 2 x y + y^{2} + 1 = 0 \tag 1$$ $$f_y = - x^{2} + 2 x y - 1 = 0 \tag 2$$
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 $$(3)$$, $$x=\pm 1$$.