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> Is there a cubic polynomial $c(x,y)$ with exactly 3 saddle point critical points?
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In other words, can a cubic polynomial in two variables have three critical points, all of which are saddle points? Or could it have the maximum 4 critical points (as per Bézout's theorem), with only one of them being a local extremum?


A broader formulation of the question could be:



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> What number of saddle points can a cubic polynomial in two variables have without having any local maximum?
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Observations:


* By [Bézout's Theorem](https://en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem) a cubic polynomial (in two variables) has at most 4 critical points.
* It is [easy to see](https://math.stackexchange.com/q/4620638/1134951) that a cubic polynomial has at most one maximum / minimum. Thus every polynomial $c(x,y)$ with 4 critical points has at least 2 saddle points. For example, the polynomial $c(x,y) = x^3-3x + y^3-3y$ has four critical points: Local maximum at $(-1,-1)$; local minimum at $(1,1)$ and saddle points at $(1,-1)$ and $(-1,1)$.
* A quartic polynomial can have all 9 critical points without having any local maximum: [Can a real quartic polynomial in two variables have more than 4 isolated local minima?](https://mathoverflow.net/q/442736/497175)
* As argued in the answer to question <https://math.stackexchange.com/q/4620663/1134951>, there is an upper bound on the number of $M-s$, where $M$ is the number local extrema (local minima and maxima) and $s$ is the number of saddle points. Here we are interested in the lower bound.



Can a cubic polynomial in two real variables have three saddle points?
The cubic $x^3 - xy^2 - 2x^2 + x$ has critical points in $(1,0)$, $(0,-1)$, $(0,1)$ and $(1/3, 0)$. The determinant of the Hessian matrix is $-4(3x^2 + y^2 - 2x)$. It assumes the values $-4$, $-4$, $-4$ and $4/3$ in these four critical points. Thus the first three of them are saddle points.


**Added later:** A simpler example is the cubic $xy(x+y-1)$ with saddle points in $(0,0)$, $(1,0)$, $(0,1)$ (arising from setting $\alpha=\beta=0$, $\gamma=1$ in the comments below).


