Is there a square with all corner points on the spiral $r=k\theta$, $0 \leq \theta \leq \infty$?

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.