How many squares with vertices on $x^2y^2 =1$ are possible? We can form a square with vertices on the curve $x^2 y^2 = 1$ and which does not intersect this curve.

Can we get any other square with vertices on $x^2 y^2 = 1$ and which does not intersect this curve? Can anyone please tell me elaborately?
 A: Yes. Rotating and scaling so as to keep the vertices on the curve, we get a square with vertices at $\pm(a,1/a)$ and $\pm(-1/a,a)$ with $a>0$. We can rotate clockwise until there is an intersection when the slope of the side from $(a,1/a)$ to $(-1/a,a)$ is less than the slope at $(a,1/a)$. This condition on $a$ may be expressed as
$$\frac{1/a-a}{a+1/a}\geq -\frac{1}{a^2},$$
or equivalently,
$$a^4 -2a^2-1\leq 0.$$
There is a similar condition for counterclockwise rotation, but rather than repeat the calculation, we can note that counterclockwise is the same as clockwise rotation followed by exchanging $x$ and $y$ coordinates (in this case $a\mapsto 1/a$). Hence, we also have the condition
$$a^{-4} -2a^{-2}-1\leq 0,$$
or equivalently,
$$a^{4} +2a^{2}-1\geq 0.$$
Solving for these two equations, we get $\sqrt{\sqrt{2}-1}\leq a\leq \sqrt{1+\sqrt{2}}$.
EDIT: We can show that these are the only such squares. Firstly suppose a square could be inscribed in the first quadrant. Let $(x_i,y_i)$ be the coordinates of the vertices with their $x$ coordinate in increasing order. Then the chords from $(x_1,y_1)$ to $(x_2,y_2)$ and from $(x_3,y_3)$ to $(x_4,y_4)$ have the same slope. By the intermediate value theorem, there are two points $c_1,c_2$ with $0<x_1<c_1<x_2<x_3<c_2<x_4$ so that the slope of the curve at those points are the same. Hence,
$$-\frac{1}{c_1^2}=-\frac{1}{c_2^2}$$
or $c_1=c_2$, which is a contradiction. Thus, at least one vertex must lie in another quadrant. The curve in the first quadrant is convex, so if at least two vertices lie in the first quadrant, then all of the vertices must lie in the convex hull of the curve in the first quadrant, as otherwise there would be an intersection with the curve. As such, at most one vertex can lie in the first quadrant, and by symmetry with the other quadrants, we conclude that each vertex must lie in a different quadrant. Now suppose we have a vertex $(a,a^{-1})$ in quadrant I, and a vertex $(b,-b^{-1})$ in quadrant II. Then there is a vertex $(b+a^{-1}+b^{-1},b-b^{-1}-a)$ in quadrant III. Since it lies on the curve in quadrant III, we have
$$(b+a^{-1}+b^{-1})(b-b^{-1}-a)=1$$
However,
$$(b+a^{-1}+b^{-1})(b-b^{-1}-a)=b^2-b^{-2}-ab+a^{-1}b-a^{-1}b^{-1}-ab^{-1}-1$$
Hence,
$$b^2-b^{-2}-ab+a^{-1}b-a^{-1}b^{-1}-ab^{-1}-2=0$$
Multiplying by $-ab^3$, we get
$$-ab^5+ab+a^2b^4-b^4+b^2+a^2b^2+2ab^3=(ab+1)(ab^3+ab-b^4+b^2)=0$$
Letting $c=ab$, we get
$$(c+1)(c(b^2+1)-b^4+b^2)=0$$
Then, either
$$c=-1\quad\text{or}\quad c=\frac{b^4-b^2}{b^2+1}$$
Now, suppose $c$ is the latter. Then if $b\leq-1$, $$a=\frac{b^3-b}{b^2+1}\leq 0,$$ which is a contradiction since $a>0$. However, if $b>-1$, then $$b+a^{-1}+b^{-1}=\frac{b^4+b^2}{b^3-b}> 0$$
which is a contradiction since $b+a^{-1}+b^{-1}<0$. Thus, we conclude that $$c=ab=-1$$
which characterizes squares with a center that coincides with the origin.
A: You can get squares with vertices on either branch by joining the points $(a,1/a)$, $(1/a,-a)$, $(-a,-1/a)$ and $(-1/a,a)$ for $a\in(0,1)$ (any other way is related to this via a rigid motion). We require the slope of the line joining $(a,1/a)$ and $(1/a,-a)$ to be less than the slope of the tangent at $(a,1/a)$. You can easily check that the condition is equivalent to
$$
a^4+2a^2-1\ge 0
$$
which is true for $a\in[\sqrt{\sqrt{2}-1},1]=[0.643594...,1]$.
You need the condition on the slopes because otherwise, by convexity of the branch in the first quadrant, the side of the square intersects the branch again.
