Comments on: Folium of Descartes

By: Joshua Zucker

Joshua Zucker — Sat, 24 Oct 2009 10:06:54 +0000

Something totally different occurred to me here. I wonder if it works?

x^3 + y^3 = (x+y)(x^2 – xy + y^2) = (x+y) ((x+y)^2 – 3xy).

I thought of this expression because it matches the 3xy on the other side.

Now let’s let s = x+y and p = xy.
We have s * (s-3p) = 3p
s^2 – 3ps = 3p
s^2 – 3ps – 3p = 0.

Hm.

Well, at least I’ve reduced a cubic to a quadratic. That has to be worth something?

Wait, it’s quadratic in s, but just linear in p!

3p(1+s) = s^2
3p = s^2/(1+s)

There, that has to help a lot.

I’m fearing that there’s still some quadratic solving to do down the road here, though.

By: Flowers for Algernon? « Continuous Everywhere but Differentiable Nowhere

Flowers for Algernon? « Continuous Everywhere but Differentiable Nowhere — Sun, 18 Oct 2009 00:36:08 +0000

[...] 18 by samjshah A couple o’ days ago, I posted a question about how to come up with a set of parametric equations equivalent to an implicit equation. It seemed to me like the general solution to this broad question would be like differential [...]

By: samjshah

samjshah — Wed, 14 Oct 2009 01:13:46 +0000

Hi everyone! Thanks for the solution and discussion. It totally works out, substituting . I haven’t had time to really think about how by doing this we are parametrizing the curve by slopes of lines through the origin. But I’m going to think a bit more about it.

@Jamie: Interesting connection! I remember reading that derivation to find all rational points on a circle.

Thanks again. I’ll report back once I really sort out in my mind what that substitution is actually doing, and can explain it to — say — a high school student.

By: Jaime

Jaime — Tue, 13 Oct 2009 20:38:14 +0000

Applying the above to the Folium of Descartes, given that the origin is an integral point, by naming z = 1+t³, and chosing only rational slopes, t=m/n, you can readily obtain integral solutions to the equation x³+y³=3xyz with the parametrization:

x = 3mn² , y = 3m²n , z = m³+n³

By: Jaime

Jaime — Tue, 13 Oct 2009 20:28:56 +0000

Hmmm… The “parametrize by slopes of lines through the origin” is a standard trick to get rational or integral points on an elliptic curve. You have to substitute it by “parametrize by rational slopes of lines through a rational point on the curve” and the origin is a rational (and integral) point of the Folium of the Descartes. A trick that comes in very handy with some Project Euler problems.

Take for instance the unit circle, x²+y²=1, and consider lines of slope t going through point (0,1), after some math you can parametrize it as:

x = -2t/(t²+1) , y =(1-t²)/(t²+1)

If you now limit yourself to rational slopes, t=-m/n, the above can be worked out to:

x = 2mn/(m²+n²), y = (m²-n²)/(m²+n²)

We are now just naming z = m²+n² away from obtaining that integral solutions of x² + y² = z² can be parametrized as

x = 2mn , y = m²-n² , z = m²+n²

which is Euclid’s parametrization of Pythagorean Triples.

By: sumidiot

sumidiot — Tue, 13 Oct 2009 18:58:57 +0000

Coincidentally, I was just reading about this yesterday. As @CalcDave indicates, you think about y=tx. Which is to say, the curve is parametrized by slopes of lines through the origin. Besides the origin, any line through the origin (and not an axis) hits the curve at exactly one point. One you set y=tx, you can plug that in the equation and solve for x and obtain the formulas you mention above.

By: CalcDave

CalcDave — Tue, 13 Oct 2009 18:43:00 +0000

I don’t know about “in general,” but if you make the substitution y = tx here, you can get that one.