Inspire by the Toeplitz Square Problem, how many squares can be drawn on the curve: $$ z(t) = e^{2\pi i\, t} + \frac{1}{\sqrt{3}} e^{2\pi i\, 3t} $$ wth $t \in [0, 2\pi]$. Here is an image:
We're up to one five nine ten squares. Here is an example that is not aligned wit the axes. can be prove there only one square here? It's not quite square, can we move it around to be a square?
E.g. Can this quadrilateral be massaged into a square? Whose points all lie on this cubic curve?
i heard the existence of one square is known for algebraic curves like this. maybe with no guarantee of exact count. a dimension count has that a quadrilateral is defined by 8 real numbers. the squares in Euclidean plane can be defined by 4 numbers. the quadrilaterals on a curve are defines by 4 numbers. generically these curves should intersect in a $$4+4-8=0$$ dimensional set. possibly an empty collection of points.
Other possible obstructions is when these curves are very bumpy. Then I think one introduces really tiny squares!
Summary - There are $13$ squares inside the curve and $9$ of them are axis-aligned.
Part I - how to locate the axis-aligned squares.
Following is a picture showing $5$ of the axis-aligned ones.
To locate the axis-aligned squares, we first define two auxillary functions for $t \in [0,2\pi]$. $$\begin{cases} X(t) &= \cos(t) + \frac{\cos(3t)}{\sqrt{3}}\\ Y(t) &= \sin(t) + \frac{\sin(3t)}{\sqrt{3}} \end{cases}$$ In terms of $X(t), Y(t)$, the original curve (in blue) is given by the parametrization $$[0,2\pi] \ni t \mapsto \gamma(t) = X(t) + iY(t) \in \mathbb{C}$$ Next, we define two auxillary curves $$ [0,2\pi] \ni t \mapsto \begin{cases} \gamma_1(t) = (X(t)+2Y(t)) + Y(t)i & \text{( light red )}\\ \gamma_2(t) = X(t) + (Y(t)+2X(t))i & \text{( light blue )} \end{cases} \in \mathbb{C} $$
In order for a pair of points $z_{\pm} = x \pm iy$ to form a vertical edge of a square, one of the two pairs $z_{\pm} - 2y$ or $z_{\pm} + 2y$ need to lie on $\gamma$. Let's say $z_{\pm} - 2y$ lies on $\gamma$, then $z_+$ lies on the slanted curve $\gamma_1$ (the light red one). We can locate axis-aligned squares centered on real axis by intersecting $\gamma$ with $\gamma_1$. By a similar argument, we can locate axis-aligned squares centered on imaginary axis by intersecting $\gamma$ with $\gamma_2$.
In first quadrant ($\Re z, \Im z > 0$) $\gamma$ and $\gamma_1$ intersect at three points. They are the points $A,B,C$ in above diagram. $\gamma$ and $\gamma_2$ also intersect in first quadrant in three points. One of them is $C$ and the other two are the points $D,E$ in above diagram.
From these $5$ points, one can construct $5$ squares whose vertices lies completely on $\gamma$. If you reflect the $4$ squares containing $A, B, D, E$ as vertices with respect to origin, one get $4$ more squares. This means there are $9$ axis aligned squares whose vertices lie on $\gamma$.
It turns out this exhausts all axis-aligned squares.
Part II - how to count the number of squares.
To count the total number of squares, we treat $p, q$ as two variables from $[0, 2\pi]$. For each $p,q$, consider the square $PUQV$ with vertices $$\begin{array}{rl} P = \gamma(p),& U = \frac{1+i}{2}\gamma(p) + \frac{1-i}{2}\gamma(q)\\ Q = \gamma(q),& V = \frac{1-i}{2}\gamma(p) + \frac{1+i}{2}\gamma(q) \end{array} $$ As functions of $(p,q)$, $P(p,q)$ and $Q(p,q)$ always lie on $\gamma$. If we can figure out the two loci in $pq$-plane for $U(p,q)$ lies on $\gamma$ and for $V(p,q)$ lies on $\gamma$, the intersection of these loci will be the parameters $(p,q)$ one need to construct a square all of its vertices lie on $\gamma$.
To achieve this, we need a simple criterion to tell whether a point $z = x+iy$ lies on $\gamma$ or not
(or at the least, a way to filter out most points that doesn't lie on $\gamma$). It turns out there is one:
Let $a = \frac{1}{\sqrt{3}}$, points on $\gamma$ are given by the parametrization:
$$z = x + iy = e^{it}(1 + a e^{2it})$$ Taking absolute value and square, we get $$z\bar{z} = 1 + 2a\cos 2t + a^2$$ Taking real part and square, we get $$\begin{align}a(z + \bar{z})^2 &= 4a(\cos t + a\cos 3t)^2 = 4a\cos\theta^2(1 + a(4\cos^2 t - 3))^2\\ &= (2a + 2a\cos 2t) ( 1 - a + 2a\cos 2t)^2\\ &= (z\bar{z} - (1-a)^2)(z\bar{z} - a(a+1))^2 \end{align} $$ Substitute $a$ back by $\frac{1}{\sqrt{3}}$ and simplify, we get $$\Lambda(z) \stackrel{def}{=} \frac{1}{\sqrt{3}}(z^2+\bar{z}^2) - (z\bar{z})^3 + 2(z\bar{z})^2 + \frac{4}{27} = 0$$ This means $\gamma$ is contained inside the hexic curve
$$\Lambda(x+iy) = \frac{2}{\sqrt{3}}(x^2 - y^2) - (x^2+y^2)^3 + 2(x^2+y^2)^2 + \frac{4}{27} = 0$$
In principle, we can locate the desired parameters by finding the two loci in $(p,q) \in [0,2\pi]$ for $$\Lambda(U(p,q)) = \Lambda\left(\frac{1+i}{2}\gamma(p) + \frac{1-i}{2}\gamma(q)\right) = 0\\ \Lambda(V(p,q)) = \Lambda\left(\frac{1-i}{2}\gamma(p) + \frac{1+i}{2}\gamma(q)\right) = 0 $$ and compute their intersections. Excluding those with $p = q$, these are the parameters for constructing the squares we seek.
Part III - the result.
To understand what the two loci look like, I wrote a program to compute $\Lambda(U(p,q))$ and $\Lambda(V(p,q))$ as some sort of heatmap. This is the heatmap I get:
Above heatmap cover the parameter space for $(p,q) \in [0,2\pi]^2$.
Outside the diagonal $p = q$, the red and blue "strips" intersect at $56$ points. Four of them, e.g. the point labelled by $X$, comes from the two self intersection points of $\gamma$, they don't give us any squares.
The remaining $52$ intersections falls into $\frac{52}{4} = 13$ groups. Each group give us one square. For each group, I have picked one of the member and label them:
In short, if I didn't make any mistake in analyzing above heatmap, there are $13$ squares formed from points on $\gamma$. $9$ of them is axis-aligned while the remaining $4$ non-axis aligned.