# rational points on the quadrifolium $(x^2 + y^2)^3 = (x^2 - y^2)^2$

I have been reading the Wikipedia page on the Quadrifolium there are two of them:

$$\begin{eqnarray*} r &=& \sin 2\theta \\ (x^2 + y^2)^3 &=& 4 x^2 y^2 \end{eqnarray*}$$

and it's $$45^\circ$$-rotated version, which is also a rose curve:

$$\begin{eqnarray*} r &=& \cos 2\theta\\ (x^2 + y^2)^3 &=& (x^2 - y^2)^2 \\ \end{eqnarray*}$$

These equations also have a $$(x,y)$$-parameterization, which can be found with some algebra.

$$\begin{eqnarray*} x &=& \cos k \theta \cos \theta = \tfrac{1}{4}\big[ e^{i(k+1)\theta} + e^{i(k-1)\theta} + e^{-i(k-1)\theta} + e^{-i(k+1)\theta} \big] \\ y &=& \cos k \theta \sin \theta \,= \tfrac{1}{4}\big[ e^{i(k+1)\theta} - e^{i(k-1)\theta} + e^{-i(k-1)\theta} - e^{-i(k+1)\theta} \big] \\ \end{eqnarray*}$$

And $$\cos k (\theta + \phi)$$ for a rotated version. Here $$\theta = 0, \frac{\pi}{4}$$ and $$k = 2$$.

Here is the picture of the quadrifolium, a kind of rose curve, and when it's rotated.  The reason I am going to look at this question again, is because I wanted to study the rational paramterization. Let $$X$$ be our curve, with a singular point at the origin $$(0,0)$$. We're trying to describe $$X(\mathbb{Q})$$.

The curve on the left right has no rational points.

Naively, if I divide one equation by the other, I get an expression for the tangent function. In fact, $$\frac{x}{y} = \tan \theta$$ for all values of $$\theta \in [0, 2\pi]$$ just as if it were a circle. Let's define a map:

$$\big[ (x,y) = (\cos \theta, \sin \theta) \big] \mapsto \Big[ ( (2x^2 - 1 )x, (1 - 2y^2) y) = ( (2 \cos^2 \theta - 1) \cos \theta , (1 - 2\sin^2 \theta)\sin \theta ) \Big]$$

And we have that $$LHS \in \mathbb{Q}$$ if $$RHS \in \mathbb{Q}$$. Is this sufficient? Is this map birational, is this enough for a paramerization?

Here is even another strategy, since we have Fourier series (even just trigonometric series because there are only 4 terms).

$$\left[ \begin{array}{c} x \\ y \end{array}\right] = \left[ \begin{array}{c} e^{3i\theta} + e^{i\theta} + e^{-i\theta} + e^{-3i\theta} \\ e^{3i\theta} - e^{i\theta} + e^{-i\theta} - e^{-3i\theta}\end{array} \right] = \left[ \begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \end{array} \right] \left[ \begin{array}{c} e^{3i\theta} \\ e^{i\theta} \\ e^{-i\theta} \\ e^{-3i\theta} \end{array} \right]$$

Or we could even try a different approach using the triple-angle formulas of trigonometry:

$$\begin{eqnarray*} \cos 3\theta &=& 4 \cos^3 \theta - 3 \cos \theta \\ \sin 3\theta &=& 3 \sin \theta - 4 \sin^3 \theta \\ \tan 3\theta &=& \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \end{eqnarray*}$$

Now these are reading as a degree-map from the circle to another algebraic variety (or scheme). In that case, why does the angle $$3\theta$$ appear in a 4-fold symmetric curve?

These singularities could be studied by Netwton's method (e.g. Puiseux series or resolution of singularities). Unfortunatley I could never get the jargon straight about the exceptional divisor and so forth. E.g. the strategy here to rationalize the lemniscate seems to amount to computing a blow-up of the curve. At this point I am going to consult an algebraic geometry textbook.

Example What is the "derivative" or tangent space at the origin $$(0,0)$$ ?

• "The curve on the left has no rational points." I can see $5$ rational points without even computing anything. – Servaes Dec 12 '18 at 17:16
• Also, I have no idea what your question is, what curves you are talking about when, or what you are doing with them. For example; "Naively, if I divide one equation by the other, I get an expression for the tangent function." Which equations do you mean? And the tangent function of what? – Servaes Dec 12 '18 at 17:19
• @Servaes Thanks... the curve in the picture on the right should have no rational points. And I am asking for a parameterization of all the rational curves on the left. E.g. for a circle $t \mapsto (\frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2})$ is a map from $\mathbb{Q}$ to the circle $x^2 + y^2 = 1$. I spend the rest of the article proposing different strategies for these rose curves. – cactus314 Dec 12 '18 at 17:26
• What rational curves on "the left"? What left? And where does this $k$ in your parametrizations come from? – Servaes Dec 12 '18 at 17:57
• (+1) interesting question. Let $G(x,y) = (x^2+y^2)^3 - (x^2-y^2)^2$. With help of a CAS, I find $G(X(s),Y(s)) = 0$ for following parameterization of the curve $$\begin{cases}X(s) &= (2-2s)M(s)\\Y(s) &= (2s-s^2)M(s)\end{cases} \quad\text{ where }\quad M(s) = \frac{(2-s^2)(2-4s+s^2)}{(2-2s+s^2)^3}$$ The derivation is a real mess, so this better stay as a comment rather than an answer ;-p – achille hui Dec 12 '18 at 18:20

## 1 Answer

You may want to find a parametrization like $$(x, y) = (f_1(t), f_2(t))$$ for some nonconstant rational functions $$f_1(t), f_2(t)\in \mathbb{Q}(t)$$. Until now, I can show you that there's no such parametrization with $$f_1(t), f_2(t)\in \mathbb{Q}[t]$$ (i.e. both are polynomial with rational coefficients).

If such parametrization exists, we should have $$x^{2} - y^{2} = f(t)^{3}$$ and $$x^{2} + y^{2} = f(t)^{2}$$ for some $$f(t)\in \mathbb{Q}[t]$$. Then $$y^{2} = \frac{1}{2}f(t)^{2}(1-f(t)),$$ so $$\frac{1}{2}(1-f(t)) = g(t)^{2}\Leftrightarrow f(t) = 1-2g(t)^{2}$$ for some $$g(t)\in \mathbb{Q}[t]$$. Then $$x^{2} = \frac{1}{2}(f(t)^{2} + f(t)^{3}) =(1-2g(t)^{2})^{2}(1-g(t)^{2})$$ Now $$\mathrm{gcd}(1-2g(t)^{2}, 1-g(t)^{2}) = 1$$, so we should have $$1-g(t)^{2} = h_{1}(t)^{2}$$ and $$1-2g(t)^{2} =h_{2}(t)^{2}$$ for some $$h_{1}, h_{2}\in \mathbb{Q}[t]$$. By the way, this give $$2h_{1}(t)^{2} -h_{2}(t)^{2} = 1$$, which is impossible if we see the first coefficients of $$h_{1}, h_{2}$$, unless they are constant. I think we can use the same argument in case of rational functions, too. (I'm not sure)