Is there a square with all corner points on the spiral $r=k\theta$, $0 \leq \theta \leq \infty$? I've posted this question on Math Stack Exchange, but I want to bring it here too, because 1) the proof seems missing in the literature, although they are some sporadic mentions and 2) maybe it requires more sophisticated topological tools to prove it or disprove it, like bordism arguments that have been used before for some cases of the square peg problem, and these are out of my reach.
The question is:

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

 A: For the non-negative reals $I= [0,\infty[$,
the Archimedean spiral is the image of the smooth injection
$$ \phi: I \to \mathbb{C},~~~ \alpha \mapsto \alpha \,e^{i\,\alpha}, $$
the inverse of which is the restriction of the complex absolute value
$$ |\,|:\mathbb{C} \to I,~~~z\mapsto|z|. $$
This implies that $z\in \mathbb{C}$ is on the spiral iff $\phi(|z|)~=~z$.
The linear bijection $R:~\mathbb{C}^2\to \mathbb{C}^2$
$$ 
\left( {\begin{array}{c}
z_1\\z_2
\end{array} } \right)
~\mapsto~
\frac{1}{\sqrt{2}}\,
\left( {\begin{array}{cc}
e^{i\,\pi/4} &e^{-i\,\pi/4}\\
e^{-i\,\pi/4}&e^{i\,\pi/4}
\end{array} } \right)\,
\left( {\begin{array}{c}
z_1\\z_2
\end{array} } \right)
$$
rotates a pair of diagonally opposite points $z_1,z_2$ of some square  by $\pi/2$ to the left into the other pair of corners of this square.
Assume that there is a square on $\phi(I)$. Then there exist a pair of reals $0\leq\alpha<\beta$  which map on   diagonally opposite corners such that with componentwise application one has
$$ 
\phi\left(\left|R \phi
\left( {\begin{array}{c}
\alpha\\\beta
\end{array} } \right)
\right| \right) ~=~
R \phi
\left( {\begin{array}{c}
\alpha\\\beta
\end{array} } \right).
$$
Calculation yields the reals that map into the other corners as
$$ 
\rho
\left( {\begin{array}{c}
\alpha\\\beta
\end{array} } \right)
~=~ \left|R \phi
\left( {\begin{array}{c}
\alpha\\\beta
\end{array} } \right)
\right|   ~=~
\frac{1}{\sqrt{2}}\, 
\left( {\begin{array}{c}
\sqrt{\alpha^2 +\beta^2 + 2\alpha\,\beta\,\sin(\beta-\alpha)}
\\
\sqrt{\alpha^2 +\beta^2 - 2\alpha\,\beta\,\sin(\beta-\alpha)}
\end{array} } \right)~=~
\left( {\begin{array}{c}
v_+\\v_-
\end{array} } \right)
\in I^2.
$$
If $\phi(v_+,v_-)$ are the other corner points of the square, one also has
$$
\rho^2
\left( {\begin{array}{c}
\alpha\\\beta
\end{array} } \right)
~=~
\rho \left( {\begin{array}{c}
v_+\\v_-
\end{array} } \right)
~=~
\left( {\begin{array}{c}
\beta\\\alpha
\end{array} } \right).
$$
Update:
Because $v_+^2+v_-^2~=~\alpha^2+\beta^2$
this is equivalent to the conditions
$$
2\,\beta^2~=~\alpha^2+\beta^2+\sin(v_--v_+)\,v_+\, v_-,
$$
$$
2\,\alpha^2~=~\alpha^2+\beta^2-\sin(v_--v_+)\,v_+\, v_-,
$$
which imply
\begin{equation}
    \beta^2-\alpha^2~=~ 2\,\sin(v_--v_+)\,v_+\, v_-.
\end{equation}
A: The points on the spiral with $\theta=$
$$
3.4535999354657\\
15.1248305526170\\
22.0370015553781\\
16.3950081067565
$$ form a square.

The numbers can be generated from this Mathematica code, if anyone wants to play with the algorithm and the initial conditions:
start = {{r, 15}, {s, 16}};    
curve[r_] := {r Cos[r], r Sin[r]};
rotate[{x_, y_}] := {-y, x};
left[p_, q_] := (p + q + rotate[q - p])/2;
right[p_, q_] := (p + q - rotate[q - p])/2;
curvedist[p_] := Norm[p - curve[Norm[p]]]^2;
test[p_, q_] := curvedist[left[p, q]] + curvedist[right[p, q]];
m = FindMinimum[test[curve[r], curve[s]], start];
{r, s, left[curve[r], curve[s]] // Norm, 
  right[curve[r], curve[s]] // Norm} /. m[[2]]

Update: Initial conditions with a Pythagorean form
start = {{r, Pi(m^2+2mn-n^2)p}, {s, Pi(-m^2+2mn+n^2)p}}

for positive integer $m,n,p$ with $m<n<m(1+\sqrt{2})$ always seem to produce squares on the spiral. The case $m=1$, $n=2$, $p=1$ leads to the solution above. Furthermore, the far point of each of these squares seems to be very close to the near point for another square! E.g. the first square ends at $\theta=22.03700...$ and the next square in that line begins with $\theta=22.03797...$

