Why does $f(x,y) = x^4 + y^4-16xy$ have critical points $(0, 0), (2, 2), (-2, -2)$?
I get the first critical point and was able to solve for $x$ all the way until $x^9/4^4$ but I do not know what to do after that. Can anyone clarify?
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The necessary condition $\nabla f = 0$ leads to the equations $x^3 = 4 y$ and $y^3 = 4x$. Inserting $y = x^3 /4$ into the second equation gives $(x^3/4)^3 = 4 x$, i.e.,
$$x^9 - 4^4 x = x(x^8 - 4^4 ) = 0.$$
The first factor becomes zero for $x = 0$ and the second factor becomes $0$ for
$$x^8 = 4^4 = (2^2)^4 = 2^8, $$
which implies $x = \pm 2$. For $x = 2$, one gets $y^3 = 8$, which implies $y = 2$. For $x = -2$, one similarly gets $y = -2$.
EDIT. A shorter way would be to notice that the symmetry of the equations $x^3 = 4y$ and $y^3 = 4x$ immediately delivers $x = y$. Therefore, $x^3 - 4x = x (x^2 - 4) = 0$, which implies $x = 0$ or $x = \pm 2$.