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Let $p$ be a prime such that $p \bmod 4 = 1$, so there exists some $i=\sqrt{-1}$ in $\mathbb{F}_p$. Furthermore, let $r \in \mathbb{N}$ be the radius of a circle such that there are $p-1$ lattice points on it. (The sums of squares function allows to compute such $r$.)

I think that for every lattice point $(x,y)$ there is a unique $z = x + iy$ in $\mathbb{F}_p$. Also, there is a generator $g$ in $\mathbb{F}_p$ such that $g^\alpha$ traverses the lattice points in the order given by their angle on the circle.

Here's an example with $g=2$:

enter image description here

(The line segments in the picture above point to $(x \bmod p)$ and $(y \bmod p)$. They describe a circle of radius $(r \bmod p)$ in the finite plane $\mathbb{F}_p \times \mathbb{F}_p$.)

Is there any literature about that phenomenon?

Can you prove the order of points fits the order given by $g^\alpha$?

Here are questions related to that topic:

Robin
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    In the picture yoou have drawn line segments from each lattice point to some other point closer to the origin. What is the significance of those line segments? – Jyrki Lahtonen May 07 '21 at 03:40
  • Those are $x \bmod p$ and $y \bmod p$. The significance is that if you're given a $z$ in $\mathbb{F}_p$ you can compute them using $i=\sqrt{-1} \pmod p$ by solving $z = x + iy$. It has the solution $x \equiv \frac{z+z^{-1}}{2}$ and $y \equiv \frac{z-z^{-1}}{2i} \pmod{p}$. This follows from trigonometry over finite fields. (See https://math.stackexchange.com/questions/334754/how-does-trigonometry-in-a-galois-field-work ) – Robin May 07 '21 at 04:11
  • (Actually, the above is slightly inaccurate. What I described was how to map a $z \in \mathbb{F}$ onto a point on the unit circle in the finite plane $\mathbb{F} \times \mathbb{F}$. However, we actually have to map $z$ onto a circle of radius $r \pmod p$ in that finite plane. IIRC the solutions become $x = \frac{z+z^{-1} r^2 }{2}$ and $y = \frac{z -z^{-1} r^2 }{2i}$ ) – Robin May 07 '21 at 04:21
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    I can sort of "explain" why the $28$ lattice points on the circle of radius $125$ are pairwise non-congruent modulo $29$. We have $i=2^7=12$. Then the ratio of relevant Gaussian primes $(2+i)/(2-i)=(2+12)/(2-12)=16$ has order seven. As $\gcd(7,4)=1, 7\cdot4=29-1$ we arrive at the decomposition $$\Bbb{F}_{29}^=\langle (2+i)/(2-i)\rangle \times \langle i\rangle$$ of multiplicative groups. It follows that the Gaussian integers $$(2+i)^k(2-i)^{6-k} i^m\qquad()$$ $0\le k\le 6$, $0\le m<4$, are pairwise non-congruent modulo $29$. But the points in $(*)$ are exactly the lattice points. – Jyrki Lahtonen May 07 '21 at 04:40
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    It is plausible, that to each prime $\equiv5\pmod8$ we can find a matching Gaussian prime $\pi=a+bi$ such that $(a+bi)/(a-bi)$ has order $(p-1)/4$ in $\Bbb{F}_p^$, when the same procedure allows us to map the elements of $\Bbb{F}_p^$ bijectively to lattice points on the circle of radius $|\pi|^{(p-5)/4}$. – Jyrki Lahtonen May 07 '21 at 04:44
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    If you already knew all of the above, then I apologize for wasting your time. I don't understand the ordering of the points on the circle yet. – Jyrki Lahtonen May 07 '21 at 04:46
  • The ordering of the points may be related to something like the argument of $(2+i)/(2-i)$ being close enough to $2\pi/7$. But that is a very fuzzy idea, bordering numerology. – Jyrki Lahtonen May 07 '21 at 05:03
  • You are totally not wasting my time! Happy to share thoughts about this since it has been circulating in my head for quite a while now :) – Robin May 07 '21 at 05:16
  • (*) makes sense to me because I came up with the radius 125 using the sums of squares function which implies that the smallest circle with $4q$ lattice points has radius $\sqrt{5^{q-1}}$. – Robin May 07 '21 at 05:19
  • Btw: I suppose, the fact that the smallest circle grows exponentially in $q$ also ensures this method does not help to solve discrete logarithms. – Robin May 07 '21 at 05:33
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    I can explain the ordering to some degree. Every solution $x^2+y^2=r^2$ spans a square with the points $(x,y), (-y,x), (-x,-y), (y,-x)$. When mapping them to the finite plane their symmetry is retained. Multiplying any of the points by $i$ (the rotation by $90^\circ$) has to yield the next point. So if $(x,y) \mapsto z = g^\alpha$ then $z \cdot i = g^{\alpha + \frac{p-1}{4}}$.

    (In simple terms: what looks like a circle here is actually $q$ different squares.)

     – Robin May 07 '21 at 05:42
  • Maybe we can extend the above argument to explain the full ordering. Here's a rough sketch: If we multiply any point on any square with some $h$ which is a generator of the subgroup of order $q$ we get a point on a different square. We can do that repeatedly to get $q$ different squares. Such an $h$ is equivalent to a rotation of $\frac{1}{q} \cdot 360^\circ$ on the big circle. – Robin May 07 '21 at 21:36
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    Let $p=13$. $5^2\equiv-1\bmod{13}$ so we can take $i=5$. We can take $r=5$ since the circle of radius $5$ has the twelve points $(5,0),(4,3),(3,4),(0,5),(-3,4),(-4,3),(-5,0),(-4,-3),(-3,-4),(0,-5),(3,-4),(4,-3)$ in order. For $x+iy$ we get, in order, $5,6,10,12,4,11,8,7,3,1,9,2$, e.g., (4,-3) gives $4+(5)(-3)=-11\equiv2\bmod{13}$. But there is no generator $g$ of the multiplicative group mod $13$ such that the powers of $g$ give the elements in that order. – Gerry Myerson May 09 '21 at 03:48
  • @GerryMyerson can you tell when it works and when it does not? – Robin May 10 '21 at 16:50
  • No. Offhand, I can't see any reason why it should ever work. For how many values of $p$ have you done the calculations to see whether it works? – Gerry Myerson May 11 '21 at 10:08

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