Here is a description of one of the constructions:
Given the two points $\color{maroon}0$ and $\color{maroon}d$ that form one side of the square:
- Construct the five $\color{gray}{\text{gray}}$ circles of common radius $\color{maroon}{od}$ shown in the diagram and locate the points $\color{maroon}a$, $\color{maroon}b$, and $\color{maroon}c$.
- Construct the two $\color{maroon}{\text{maroon}}$ circles: one with center $\color{maroon}c$ and radius $\color{maroon}{cb}$, and one with center $\color{maroon}d$ and radius $\color{maroon}{ad}$. Note that $\color{maroon}{ad}$ and $\color{maroon}{cb}$ have the same length.
- Locate the point of intersection $\color{pink}e$ of the two maroon circles.
- Construct the $\color{pink}{\text{pink}}$ circles of radius $\color{pink}{oe}$ at centers $\color{maroon}c$ and $\color{maroon}d$.
- The point of intersection, $\color{pink}f$, of the pink circles is a vertex of the square.
- Draw the $\color{darkgreen}{darkgreen}$ circle centered at $\color{pink}f$ of radius $\color{maroon}{od}$.
The point of intersection, $\color{darkgreen}g$, of the darkgreen circle with the gray circle centered at $\color{maroon}d$ is the final vertex of the square.
Justification of step 5:
From step 1., $\color{maroon}{aboc}$ is a rhombus with common side length $ l(\color{maroon}{co})$. Since the point $\color{pink}e$ is equidistant to the points $\color{maroon}c$ and $\color{maroon}d$ , the segment $\color{pink}e\color{maroon}o$ is perpendicular to the segment $\color{maroon}{cd}$. Since $\color{pink}f$ is equidistant to the points $\color{maroon}c$ and $\color{maroon}d$, the points $\color{pink}e$, $\color{pink}f$, and $\color{maroon}o$ are colinear and the segment $of$ is perpendicular to the segment $\color{maroon}od$.
We need to show that $l(fo)=l(co)$. Proceeding with some abuse of notation:
Considering the rhombus $aboc$, we have $ cb^2+ao^2=2(ac^2+co^2 ); $ or, since $ao=co=ac$, $ cb^2=3co^2. $ Since $cb=ce$, we have $\tag{1} ce^2=3co^2 $
Considering now the right triangle $ceo$: $\tag{2} ce^2=eo^2+co^2. $ Combining equations $(1)$ and $(2)$: $\tag{3} eo^2=2co^2. $
Considering now the right triangle $cfo$: $\tag{4} cf^2=fo^2+co^2 $ since $cf=oe$: $\tag{5} oe^2=fo^2+co^2 $ From $(3)$ and $(5)$ now, we finally obtain $ fo=co, $ as desired.