A more usual definition of a double point on a curve is that it is a point where the curve crosses itself. For a more precise definition, see here. On that same Wikipedia page, there is a graph of the folium of Descartes. That is a cubic curve, with the equation , which has a double point at the origin. The unit circle cuts the folium of Descartes at four points. So if you multiply their equations together, you get a quintic curve whose graph is the superposition of the graphs of the folium and the circle, and therefore has five double points.
I don't know whether a curve with a quintic equation can have six double points.
Hi Opalg!
I thank you very much for your intelligent solution
about the five double points.
At last I have a quintic five double points!
Perhaps you help me again for the following problems:
May you create a quintic five double points irreducible?
It is no hope for a quintic six double point?
I am not knowledgeable about algebraic geometry and I cannot answer those questions. The best information I could find on the web is a 1902 Cornell University PhD dissertation by Peter Field, which you can find here. Click on "Page 1" to get a pdf file of the first chapter, which seems to state that there are many irreducible quintic curves with six double points. The text is illustrated with hand-drawn diagrams of such curves (look at Plate I at the end of that chapter), but I could not find any explicit equations for them.
It seems that double points are known to those in the trade as "crunodes". If you have access to a university library with old books on algebraic geometry (and an Italian university ought to have lots of those ) then look for "quintic" and "crunodes" in the index. You might get lucky.
Hello Opalg
I thank you for your useful information…
I wanted a your opinion on a method, than it has sprung to mind,
in order to find the a 6-double points quintic equation...
Practically, fixed in the cartesian plan 8 points, 6 double and 2 simple.
if F(x,y) is quintic polynomial and Fx(x,y) and Fy(x,y) its derivatives,
then for every Xi,Yi of the 6 double points I have the three equations
F(Xi,Yi)=0, Fx(Xi,Yi)=0, Fy(Xi,Yi)=0
and for every Xj,Yj of the 2 simple points I have the equation
F(Xj,Yj)=0.
I have so a linear omogeneous system of 20 equations in the 20 unknown quantities
that are the essential coefficients of the quintic, resolved which I have the equation…
That you think some?