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I'm interested in the number of solutions $x^2＋y^2＝1$ in $\Bbb{F}_q$. I searched this site also, and by combining the information obtained there, I made the following answer.

The curve $x^2+y^2=1$ has genus zero, so its projective version $X^2+Y^2+Z^2=0$ always has exactly $q+1$ points in $\Bbb{P}^2(\Bbb{F}_q)$. Two of those points will lie on the line at infinity when $-1$ has a square root in $\Bbb{F}_q$.So,when $q≡1mod4$, $q$ー1 is the answer, and when $q≡3mod4$, $q＋1$ is the answer.

I have two questions.

① Why genus zero deduces the curve has $q＋1$ points?

②Why infinity consists of two points?

Moishe Kohan
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Pont
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    Hasse-Weil bound does not leave much room when $g=0$. – Jyrki Lahtonen Sep 10 '20 at 20:38
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    On the line of infinity we have $Z=0$, so (intersecting with the curve) $X^2+Y^2=0$. This has two points $[1:i:0]$ and $[1:-i:0]$ on the curve when $i$ exists and no points otherwise. – Jyrki Lahtonen Sep 10 '20 at 20:40
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    Viewed differently, genus zero means that the curve has a rational parametrization, so $x=x(t)$, $y=y(t)$ for some $x(t),y(t)\in\Bbb{F}_q(t)$, $t$ transcendental over $\Bbb{F}_q$. Not unlike the rational parametrization of the real unit circle: $x=(1-t^2)/(1+t^2)$, $y=2t/(1+t^2)$ we learned about when studying integration of trig functions (or studied Pythagorean triples). – Jyrki Lahtonen Sep 10 '20 at 20:43
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    That parametrization actually works over $\Bbb{F}_q$ as well, and you see that at $t=\pm i$ the point $(x,y)$ lands on the line at infinity. – Jyrki Lahtonen Sep 10 '20 at 20:46
  • My guess is that this question has been handled on the site already. Alas, I need some shut-eye, and cannot search now. Hope you can make some progress. – Jyrki Lahtonen Sep 10 '20 at 20:48
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    @JyrkiLahtonen See this: https://math.stackexchange.com/questions/3201863/how-many-solutions-of-the-equation-ax2-by2-1-are-there-with-x-y-%e2%88%88-ma/3202865#3202865 – Alex Youcis Sep 10 '20 at 20:53

1 Answers1

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Note that your answer doesn't address what happens when $q$ is even. This case degenerates because $x^2 + y^2 = (x + y)^2$ so the equation reduces to $x + y = 1$ which, of course, has $q$ solutions.

Q1: Abstractly, a smooth projective curve of genus $0$ with a rational point is isomorphic to $\mathbb{P}^1$. Concretely, in this case we can just write down such an isomorphism: it's given by

$$(A : B) \mapsto (A^2 - B^2 : 2AB : A^2 + B^2)$$

which is just the familiar formula for Pythagorean triples, and is the homogeneous version of the map given by Jyrki in the comments. It's a good exercise to see how this map arises by intersecting the curve $C$ with a line passing through $(1 : 0 : 1)$ in $\mathbb{P}^2$ (visualizing the real affine case of a line passing through $(1, 0)$ intersecting a circle may help).

Q2: As Jyrki says in the comments, the points at infinity are the points satisfying $X^2 + Y^2 = 0$. If $q \equiv 1 \bmod 4$ or $q$ is even then $-1$ has a square root so there are two points at infinity $(1 : \pm i : 0)$. If $q \equiv 3 \bmod 4$ then $-1$ is not a square root so there are no points at infinity. And if $q$ is even then $X^2 + Y^2 = (X + Y)^2$ so there is one point at infinity $(1 : 1 : 0)$.

An alternate way to produce this count is to write $x^2 + y^2$ as the norm $N(x + iy)$ where $x + iy \in \mathbb{F}_q[i]/(i^2 + 1)$. The isomorphism type of this quotient depends on how $t^2 + 1$ factors over $\mathbb{F}_q[t]$.

If $q \equiv 3 \bmod 4$ then you can check that $t^2 + 1$ is irreducible so $\mathbb{F}_q[i] \cong \mathbb{F}_{q^2}$. You can also check that the norm $N : \mathbb{F}_{q^2} \to \mathbb{F}_q$ is surjective, so its kernel has size $\frac{q^2 - 1}{q - 1} = q + 1$.

If $q \equiv 1 \bmod 4$ you can check that $t^2 + 1$ is reducible so $\mathbb{F}_q[i] \cong \mathbb{F}_q^2$. The norm is again surjective, so its kernel has size $\frac{(q - 1)^2}{q - 1} = q - 1$.

If $q$ is even $t^2 + 1 = (t + 1)^2$ so $\mathbb{F}_q[i] \cong \mathbb{F}_q[\varepsilon]/\varepsilon^2$ has nilpotents. The norm is again surjective, so its kernel has size $\frac{q(q - 1)}{q - 1} = q$.

Qiaochu Yuan
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    You should probably mention that every smooth conic section has a rational point over a finite field so the fact that smooth proper connected curves with a rational point being rational applies always over $\mathbb{F}_q$. This is related to the fact that $\mathrm{Br}(\mathbb{F}_q)=0$. – Alex Youcis Sep 10 '20 at 21:10
  • Right, of course, this follows from Chevalley-Warning (https://en.wikipedia.org/wiki/Chevalley%E2%80%93Warning_theorem). My number theory is slightly rusty! – Qiaochu Yuan Sep 10 '20 at 21:12
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    That's right although it's simpler here. An application of the Riemann-Roch theorem shows that smooth (geometrically) connected genus 0 curve over any field $k$ is isomorphic to a conic section of the form $V(ax^2+by^2-z^2)\subseteq \mathbb{P}^2_k$. If $k$ is finite one can count the size of the sets ${ax^2}$ and ${1-by^2}$ to deduce that they have a point in common and thus $V(ax^2+by^2-z^2)$ has a rational point even on the affine patch $V(ax^2+by^2-1)\subseteq \mathbb{A}^2_k$. – Alex Youcis Sep 10 '20 at 21:15
  • Thank you ! I completely understood question② and your second approach, but I have still confused with ①. I understood the rational point of the curve is isomorphic to P1, but why the number of rational points is q＋１？　P1 has q^2-1 elements... – Pont Sep 10 '20 at 21:45
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    @bellow: $\mathbb{P}^1$ has $\frac{q^2 - 1}{q - 1} = q + 1$ elements. There are the "affine" points $(x : 1), x \in \mathbb{F}_q$ and then the point at infinity $(0 : 1)$. – Qiaochu Yuan Sep 10 '20 at 22:09