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By Bézout's theorem a quartic polynomial $p(x,y)$ can have at most 9 isolated critical points. Can all of them be saddle points?

In case of a cubic polynomial there is a mechanical way to answer this type of questions: One can find a general form of such polynomials with critical points at three given points, e.g. at $(0,0),(0,1),$ and $(1,0)$, and then play with the remaining free parameters until a polynomial with the desired properties if found, as there is only one remaining critical point whose position was not normalized by affine transformation [1], [2].

However, the above approach won't work in case of a quartic polynomial, as then affine transformation normalizes position of only 3 out of the 9 potential critical points.

Note that a quartic polynomial in two real variables can have at most 5 minima out of its 9 potential critical points [3], [4]. Can it reversely have no extreme points, so that all 9 of the potential critical points would be saddle points?

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By (3.1) of Counting Critical Points of Real Polynomials in Two Variables by Alan Durfee, Nathan Kronefeld, Heidi Munson, Jeff Roy, Ina Westby a degree $d$ polynomial with only nondegenerate critical points can contain at most $d(d-1)/2$ saddle points. For $d=4$ this gives a max of $6$ saddle points.

This bound is attained by a product of $d$ linear polynomials, as long as each pair of lines intersects at a different point, as then the $d(d-1)/2$ intersection points are all saddle points.

(This paper was cited in a paper cited in one of the answers you cite.)

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    $\begingroup$ Thanks a lot for the answer, I finally read the paper, it is very nicely written with a lot of intuition. $\endgroup$
    – Pavel Kocourek
    Commented May 17, 2023 at 17:44

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