WHAT GENERAL RESULTS MIGHT THERE BE?
What can be said about the critical points of real biquadratics of the form
a00+(a10*x+a01*y)+(a20*x^2+a02*y^2)+a11*x*y+x*y*(a21*x+a12*y)+a22*x^2*y^2
in which a22 is nonzero and we know the sign of a22 ?
One can also consider the curve defined by the above being 0. Then genus, singular points, ...?
The Surface: z =f(x,y) for a biquadratic is a genus 1 landscape.
The Stationary Points: Found via a quintic in x (or lower degree in special cases, e.g. Lyness gives cubic)
Symmetry: (a10=a01, etc.) The quintic factorises into a quadratic (off-diagonal pairs) and a cubic (points on y=x).
How many ovals: Never more than 2. Lyness can have none, Edwards, etc, has one.
The "Two Oval" Recipe: Requires the quadratic discriminant to be positive (creating two "wells" or "peaks") and the level c to be on the correct side of the saddle point value.
The a22 Role: Dictates global boundedness; its sign relative to a00 and the discriminants determine if those ovals are "nested" or if the curve is even real. In the fertiliser example we have a00>0 and a22<0.
Bézout's theorem refers only to the case of two variables, and asserts that, if two plane algebraic curves of degrees d1 and d2 have no component in common, they have d1*d2 intersection points, counted with their multiplicity, and including points at infinity and points with complex coordinates.
Applying this in the case a22 nonzero, we have the x and y derivatives of total degree 3, so d1=3=d2. We get 5 in finite plane or complex and 4 "at infinity".
MISCELLANEOUS APPLICATIONS
Elliptic curve cryptography, finite fields. E.g Edwards curve
QRT maps
Poncelet Porism. 1-1 correspondence between positions of bicentric n-gons and points on a biquadratic curve
4-bar linkages. 1-1 correspondence between positions of a 4 bar linkage and points on a biquadratic curve
W. Wu: Evans Triangles and Edwards Curves
V. Dragovic and M. Radnovic: Finite Groups of Random Walks in the Quarter Plane and Periodic 4-bar Links
VARIOUS TYPES OF 4TH-TOTAL-DEGREE BIQUADRATICS
Lyness x y (x y +1) (a+b) - a b (a b+1) (x+y) ==0 this haa a00=0
Edwards
McMillan maps: general symmetric biquadratic
A BIT MORE ON FEB 2026 MATHEMATICA
Mgen = {{am00, am01, am02}, {am10, am11, am12}, {am20, am21, am22}};
biquadgen = ({{1, x, x^2}} . Mgen . Map[List, {1, y, y^2}])[[1, 1]];
(* A short calculation yields the quintic *)
critPtsEq = (4*am02^2*am10 - 2*am01*am02*am11 + am01^2*am12) +
2*(-(am02*am11^2) + 4*am02*am10*am12 + 4*am02^2*am20 - 2*am01*am02*am21 + am01^2*am22)* x +
(-(am11^2*am12) + 4*am10*am12^2 + 16*am02*am12*am20 - 6*am02*am11*am21 -
2*am01*am12*am21 + 8*am02*am10*am22 + 2*am01*am11*am22)*x^2 +
4*(2*am12^2*am20 - am11*am12*am21 - am02*am21^2 + 2*am10*am12*am22 + 4*am02*am20*am22)*x^3 +
(-3*am12*am21^2 + 16*am12*am20*am22 - 2*am11*am21*am22 + 4*am10*am22^2)*x^4 +
2*am22*(-am21^2 + 4*am20*am22)*x^5
(* the quintic critPtsEq factors in the symmetric case. *)
symu[u_] := u /. {am10 -> am01, am20 -> am02, am21 -> am12};
Mgensym = symu[Mgen]
Factor[symu[critPtsEq] ]
(* gives product of quadratic and cubic below *)
critPtsEq3sym = (am01 + 2 am02 x + am11 x + 3 am12 x^2 + 2 am22 x^3);
critPtsEq2sym = 4*am02^2 - 2*am02*am11 + am01*am12 +
(2*am02*am12 - am11*am12 + 2*am01*am22)* x +
(-am12^2 + 4*am02*am22)*x^2;
Assuming symmetry in the N-optimisation and using a0>0, a1>0, a2<0 one might write the biquadratic profitr
bxr = -ar02*(n0 + ar01/(2*ar02) + r)*(r - n0 - ar01/(2*ar02));
bfr = bxr*(bxr /. n0 -> nx);
profitr = bfr - cn*(n0 + nx)
Mrprofit = Map[Simplify, CoefficientList[profitr, {n0, nx}]]
dprofitr0 = D[profitr, n0]; dprofitrx = D[profitr, nx];
critPtnEq = Numerator[Factor[dprofitr0 /. ( Solve[dprofitrx == 0, nx][[1]])]]
critPtnEq2= (ar01*cn + ar01^2*ar02*r^2 - 4*ar02^3*r^4 +
2*ar02*n0*(cn + 2*ar01*ar02*r^2)+ 4*ar02^3*n0^2*r^2;
critPtnEq3 = (-ar01^3 + 4*ar02*cn+4*ar01*ar02^2*r^2 )-2*ar02*n0*(3*ar01^2 - 4*ar02^2*r^2) -
12*ar01*ar02^2*n0^2 - 8*ar02^3*n0^3 ;
maxProfitn0 = -(ar01/(2 ar02)) - 2 *r*Sin[1/3 ArcSin[(3 Sqrt[3] *cn)/(4*ar02^2*r^(3))]]/Sqrt[3];
check = FullSimplify[critPtnEq3 /. {n0 -> maxProfitn0}]
(* 0, checks *)
Use hessian to show it is a max, and the other 4 finite plane critical points are all saddles.