<s>[INST] Are there any other dyadic points on the unit circle? There are four points on the unit circle with integer coordinates
$$
(\pm 1,0)\;, \; (0,\pm 1) 
$$
Besides these four, are there any other points on the unit circle with coordinates from $ \mathbb{Z}[1/2] $?
If we extend the coordinate ring to $ \mathbb{Z}[1/\sqrt{2}] $ then we get the eight points on the unit circle
$$
(\pm 1,0)\;, \; (0,\pm 1) \;, \; (\pm 1/\sqrt{2},\pm 1/\sqrt{2}) 
$$
Besides these eight, Are there any other points on the unit circle with coordinates from $ \mathbb{Z}[1/\sqrt{2}] $?
 [/INST] Using suggestion from wasn't me we can prove that these are the only four dyadic points on the unit circle. Let $ (x,y) $ be dyadic point on the unit circle then we can write
$$
x=\frac{a}{2^k} \; y=\frac{b}{2^\ell}
$$
for $ a,b \in \mathbb{Z} $ and $ k, \ell $ nonnegative integers. So we must have
$$
\frac{a^2}{2^{2k}}+\frac{b^2}{2^{2\ell}}=1
$$
Without a loss of generality assume $ k \geq \ell $. Then we must have
$$
a^2+(2^{k-\ell}b)^2=2^{2k}
$$
The only solution to this is $ a= \pm 2^k, 2^{k-\ell}b=0 $ or $ a= 0, 2^{k-\ell}b=\pm 2^k $. For a proof of this fact see
Prove there are 4 integer roots on $x^2+y^2=2^n$
So we can conclude that either $ x=\frac{a}{2^k}=\pm 1, y=\frac{b}{2^\ell}=0 $ or $ x=\frac{a}{2^k}=0, y=\frac{b}{2^\ell}=\pm 1 $.
For the second question let
$$
(x,y)=(a+\frac{b}{\sqrt{2}},c+\frac{d}{\sqrt{2}})
$$
be a $ \mathbb{Z}[1/\sqrt{2}] $ point on the unit circle. Then
\begin{align*}
x^2+y^2&=1\\
(a+\frac{b}{\sqrt{2}})^2+(c+\frac{d}{\sqrt{2}})^2&=1\\
a^2+\frac{b^2}{2}+\sqrt{2}ab+c^2+\frac{d^2}{2}+\sqrt{2}cd &=1\\
\sqrt{2}(ab+cd) &=1-(a^2+\frac{b^2}{2}+c^2+\frac{d^2}{2})\\
\end{align*}
The right hand side is an integer. In order for the left hand side to be an integer we must have $ ab+cd=0 $. Which implies
$$
1=a^2+\frac{b^2}{2}+c^2+\frac{d^2}{2}
$$
This is only possible if $ a=\pm 1 $ and everything else is $ 0 $ (2 cases) or $ c=\pm 1 $ and everything else is $ 0 $ (2 cases) or $ a=b=0 $ and $ a=\pm1, b=\pm 1 $ (4 cases). That exactly restricts us to the eight $ \mathbb{Z}[1/\sqrt{2}] $ points on the unit circle listed in the question.
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