Is there a quartic polynomial in two variables that have multiple local minima and no other critical points? 
Question. Can a degree 4 polynomial $p:\mathbb R^2 \to \mathbb R$ have $N\geq 2$ local minima and no other critical points?

I got this question when trying to answer: How many strict local minima a quartic polynomial in two variables might have?
A closely related question asks: If a two variable smooth function has two global minima, will it necessarily have a third critical point? Among the answers @RiverLi gives an example of the function $$
f(x,y)=(x^2-1)^2+(x^2y-x-1)^2, \tag{*}\label{RL}
$$
which is a polynomial, but of degree 6.
Notice that polynomials like \eqref{RL}  can be written as $p(x,y)=u(x,y)^2 + v(x,y)^2$. Since we need $p$ to be degree 4 polynomial, $u$ and $v$ must be quadratic polynomials. We need there to be a set $P\subset \Bbb R^2, |P|=N$, of points such that:

*

*$(x,y)\in P \ \ \Longleftrightarrow \ \ u(x,y)=v(x,y)=0$;

*$(x,y)\in P \ \ \Longleftrightarrow \ \ u_x u + v_x v = u_y u + v_y v =0$ when evaluated at $(x,y)$.


For what values of $N\geq 2$ is there a set $P\subset \Bbb R^2, |P|=N$ and quadratic polynomials $u,v:\Bbb R^2 \to \Bbb R$ satisfying conditions 1 and 2?

My guess is that for $N=2$ at least one of the polynomials $u,v$ has to be of the third order as in \eqref{RL}. However, it is possible that there would be desirable $u,v$ for $N=3$ or $N=4$. Note that by Bézout's theorem condition 1 can not be satisfied at more than 4 points given that $u,v$ are quadratic.

Alternatively, is there another form than $p=u^2+v^2$ that could represent a quartic polynomial $p$ with $N\geq 2$ local minima and no other critical points?


Disclaimer. I asked about the special case of $N=2$ on https://math.codidact.com.
 A: This paper: "Counting Critical Points of Real Polynomials in Two Variables":
http://www.jstor.org/stable/2324459
gives a partial answer for the case $N=4$, which they prove is not possible.
The paper defines the global index of a polynomial as the "turning number" of its gradient vector field, see:
https://en.wikipedia.org/wiki/Vector_field#:~:text=The%20index%20of%20a%20vector,a%20source%20or%20sink%20singularity
and
https://www.researchgate.net/figure/A-critical-point-of-a-normalized-vector-field-is-a-point-where-its-direction-is-undefined_fig5_358687586.
They give the formula for the global index: $$i=m+n-s$$ where $i$ is the index, $m$ is the number of local maxima, $n$ is the number of local minima, and $s$ is the number of saddle points counted with multiplicity.
It further states that the local index of an isolated non-degenerate maximum or minimum can be at most $1$, so isolated non-degenerate maximum and minimum doesn't have any index multiplicity for a polynomial.
In the paper, proposition 2.5 states that the index $i$ of a polynomial $f(x,y)$ of degree $d$ about a circle $C$ in the plane is bounded according to $$|i| \leq d-1$$
If we take the circle large enough it will enclose all critical points and give us the global index.
If we insert $d=4$ for a degree $4$ polymomial we get that the index is bounded as:
$$|i| \leq 3$$
Moreover, if we only have local minima ($m=0$, $s=0$) then we have:
$$i=n$$
Now, for $n=N=4$ we get $i=4>3$,  so that $N=4$ is not possible.
Later on, the authors comment that they don't believe $i=d-1$ is possible which would also exclude the case $N=3$ for $d=4$, however this is only a conjecture. It would be very nice if someone could find a counter example to this. Very little seems to be written on the cases $N=3$ and $N=2$.
