Is there a square with all corner points on the spiral $r=k\theta$, $0 \leq \theta \leq \infty$? While reading about the square peg problem, I found this paper of Jerrard, where he described that for the spiral
$$r = k\theta \quad 2\pi \leq \theta \leq 4\pi $$
if we join the endpoints, you can only draw one square that all of its corner points lie in this new closed curve. Indeed, he remarked that "The square has one corner point on the straight line segment, and does not lie entirely in the interior".
The problem is that, in a way, he said that the spiral alone (open curve) doesn't contain any inscribed square. I haven't seen a proof of this in the literature, so the question is:

Is there a square that all of its corner points lie in the spiral
$$r = k\theta \quad 0 \leq \theta \leq \infty $$
?


Update: Users have shown that it's possible to draw "quasi"-squares in the spiral. This support the idea that a square should exist and that an approach using a system of equations and a fixed point theorem could be the solution. I'm wondering, what parameters (polar coordinates, @sirous parameters) should be used to write the equations for the side/diagonal lengths in the most clear way? Let's try!
Let $A$, $B$, $C$, $D$ the points of our desired square in the spiral and without losing generality, we take $k=1$. Using the cosine theorem:
$$d_{A,B}^2=\theta_{A}^2+\theta_{B}^2-2\theta_{A}\theta_{B}\cos(\theta_{A}-\theta_{B})$$
$$d_{B,C}^2=\theta_{B}^2+\theta_{C}^2-2\theta_{B}\theta_{C}\cos(\theta_{B}-\theta_{C})$$
$$d_{C,D}^2=\theta_{C}^2+\theta_{D}^2-2\theta_{C}\theta_{D}\cos(\theta_{C}-\theta_{D})$$
$$d_{D,A}^2=\theta_{D}^2+\theta_{A}^2-2\theta_{D}\theta_{A}\cos(\theta_{D}-\theta_{A})$$
where $\theta_{X}$ is the angle that define the point $X$ in the spiral and $d_{center,X} = \theta_{X}$ by the polar equation of the spiral.
This gives 3 independent equations:
$$d_{A,B}^2=d_{B,C}^2$$
$$d_{B,C}^2=d_{C,D}^2$$
$$d_{C,D}^2=d_{D,A}^2$$
and we need to add a fourth one to discriminate between a square and a rhombus. This could be a right angle between the segment AB and CD or as @Intelligenti pauca said, 2 times the length of the side squared should be equal to the length of the diagonal squared.
$$2d_{A,B}^2=d_{A,C}^2$$
This should give 4 equations and 4 unknowns ($\theta_{A},\theta_{B},\theta_{C},\theta_{D}$), exciting!, but the system is non-linear, so we can't establish if it has a solution a priori
PD: I created this Geogebra graph, so you can slide the points along the spiral and try new ideas!
 A: 
Comment:
In this figure the pitch of spiral is $6.4$mm and the radius of circle the spiral constructed on is $32$mm(this is not important). This figure shows that the coordinates of vertices of square is the solution of the following system of equations:
$$\begin{cases}r=k\theta; 0<\theta<\infty, center  (0, 0)\\ R=m\alpha ; \alpha=[0, 2\pi], center  (130.4, 13.9) \end{cases}$$
where r is for spiral, R is the radius of circle. Another equation must be included in this system which has pitch (p)as a parameter. For particular values of r, R and p we may have a square with vertices on spiral.This can be seen when we compare small square with the big one.
Update: I explain how I drew the figure:
The spiral is constructed on a circle radius $r_1$.
1- draw a spiral with arbitrary pitch p .
2- take an arbitrary point O on spiral  as the center of the square. Draw a square and it's inscribed circle, with another square in, for comparison.
3- fix a corner of the circle A on spiral for rotation of circle about.
4- Rotate the square until the extension of the side AB containing the point A  passes the center of the circle the spiral based on(the center of spiral. so $AB=n. p$. If AB has fixed measure then we alter the value of pitch. I played with measures of AB and p such that other two corners located on spiral.
