8
$\begingroup$

Consider the finite field $\mathbb{F}_q$. Schur (1916) proved that, given $n$, when the field is sufficient large, this equation,

$$x^n+y^n= z^n$$

always has a nontrivial solution.

What conditions does the number of solutions satisfy?

  • I. Schur, Über die Kongruenz $x^{m} + y^{m} \equiv z^{m} \pmod{p}$, Jahresber. Deutschen Math. Verein. 25 (1916), 114–117.
$\endgroup$
2
  • 9
    $\begingroup$ It's hard for me to make sense of "what the conditions the number of the solutions satisfy", even at the level of syntax. $\endgroup$
    – YCor
    Jul 14, 2017 at 8:25
  • 2
    $\begingroup$ Lang-Weil tells you that the number $f(q)$ of solutions in $\mathbf{F}_q$ is $q^2+O(q^{3/2})$. Of course one can write $f(q)=1+(q-1)g(q)$, and equivalently $g(q)=q+O(\sqrt{q})$; here $g(q)$ is the number of points of some smooth irreducible projective curve. $\endgroup$
    – YCor
    Jul 14, 2017 at 8:29

2 Answers 2

13
$\begingroup$

There are some results and references in Lang, Cyclotomic Fields, 1.§6 (p. 22ff. in Cyclotomic Fields I and II, Combined Second Edition).

Let $V(d)$ be the Fermat curve of degree $d$. Theorem 6.1. The number of points of $V(d)$ (in affine space) is $q^2 - (q-1)\sum\chi^{a+b}(-1)J(\chi^a,\chi^b)$, the sum over integers $a,b$ with $0 < a,b < d$ and $a+b \not\equiv 0 \pmod{d}$ and $\chi$ the character such that $\chi(u) = \omega(u)^{(q-1)/d}$ with $\omega: \mathbf{F}_q \to \mu_{q-1}$ the Teichmüller character and the Jacobi sum $J(\chi_1,\chi_2) = -\frac{S(\chi_1)S(\chi_2)}{S(\chi_1\chi_2)}$ and the Gauß sum $S(\chi) = \sum_u\chi(u)\lambda(u)$ and $\lambda: \mathbf{F}_q \to \mu_p, \lambda(x) = \exp(2\pi i/p\mathrm{Tr}(x))$.

$\endgroup$
9
$\begingroup$

It is more or less easy to obtain, via exponential sums, an asymptotic formula for the number $J$ of solutions of the congruence

$$x^{n}+y^{n} \equiv z^{n} \pmod{p}$$

where $$1 \leq x, y, z \leq p-1.$$

Indeed, since

\begin{eqnarray*} J &=& \sum_{x=1}^{p-1} \sum_{y=1}^{p-1} \sum_{z=1}^{p-1} \frac{1}{p}\sum_{a=0}^{p-1}e^{2 \pi i \frac{a(x^{n}+y^{n}-z^{n})}{p}}\\ &=& \frac{(p-1)^{3}}{p}+ \frac{1}{p} \sum_{a=1}^{p-1} \sum_{x=1}^{p-1}\sum_{y=1}^{p-1} \sum_{z=1}^{p-1}e^{2\pi i \frac{a(x^{n}+y^{n}-z^{n})}{p}}\\ &=& \frac{(p-1)^{3}}{p} + \frac{1}{p}\sum_{a=1}^{p-1}\left|\sum_{x=1}^{p-1}e^{2\pi i \frac{ax^{n}}{p}}\right|^{2}\sum_{y=1}^{p-1}e^{2\pi i \frac{ay^{n}}{p}}\\ \end{eqnarray*}

and

$$ \sum_{a=0}^{p-1} \left| \sum_{x=1}^{p-1} e^{2\pi i \frac{ax^{n}}{p}} \right|^{2} \leq n\,p\,(p-1)$$

and

$$\left|\sum_{y=1}^{p-1}e^{2\pi i \frac{\alpha y^{n}}{p}}\right| \leq n\sqrt{p}$$

for any integer $\alpha$ which is not divisible by $p$, it follows that

$$J = \frac{(p-1)^{3}}{p}+O\left(n^{2}(p-1)\sqrt{p}\right).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.