21
$\begingroup$

If p is a prime other than 7, can every integer be written as sum of two cubes modulo p? Has Waring's problem mod p for cubes been proved simply and directly? Thanks for your proof. Lemi

$\endgroup$

4 Answers 4

28
$\begingroup$

You ask if $X^3+Y^3=a$ has a solution in $\mathbb F_p$ for each $a\in\mathbb F_p$. That's clear if $a=0$. If $a\ne0$, then $X^3+Y^3=a$ describes a non-singular cubic curve, so by Hasse's Theorem, its projective completion has at least $p+1-2\sqrt{p}\;$ $\mathbb F_p$-points. There are at most $3$ points at infinity, so you are done as $p+1-2\sqrt{p}-3>0$ for $p>7$.

I don't think that there is a simpler proof, except for the trivial case $p\not\equiv1\pmod{3}$. There are slightly weaker (and easier to prove) versions of Hasse's Theorem, which are good enough for your purpose.

A nice account of Hasse's Theorem (in characteristic $\ne2,3$) is in this paper by Chahal and Osserman. Another proof of a weaker version of Hasse's Theorem is in Section 1.6 in the book Logarithmic Forms and Diophantine Geometry by Baker and Wüstholz.

Added later: Taking up the suggestions by Gjergji Zaimi and Lucia, the result follows from Vosper's addition to the Cauchy-Davenport Theorem once we know that the cubes in $\mathbb F_p^\star$ for $p\equiv1\pmod{3}$ don't form an arithmetic progression. That, however, is easy to see: The sum over the cubes vanishes, and so does the sum over the squares of the cubes if $p>7$. Assume that there are $u,v$ such that $\sum_{i=1}^{(p-1)/3}(u+iv)^m=0$ for $m=1$ and $m=2$. For $m=1$ we get $3u+v=0$, and $m=2$ yields $3^3u^2+2\cdot3^2uv+v^2=0$, hence $0=2\cdot3^2\cdot11u^2$, and therefore $u=v=0$.

$\endgroup$
3
  • 9
    $\begingroup$ In this case Hasse's theorem has a simpler proof: if one computes the number of points using additive characters you get $p-2$ plus the contribution from the Gauss sums attached to the two characters of order exactly $3$. $\endgroup$
    – Lucia
    Oct 24, 2014 at 11:12
  • 8
    $\begingroup$ There should be a simpler proof using the Cauchy Davenport inequality. (We might need Vosper's result which also characterizes when equality can happen) $\endgroup$ Oct 24, 2014 at 11:14
  • 7
    $\begingroup$ @GjergjiZaimi: That's a nice idea (using that if $n$ is missed then so are all numbers of the form $nr^3$)! One would have to check that the cubes don't form an arithmetic progression, which indeed happens for $p=7$. $\endgroup$
    – Lucia
    Oct 24, 2014 at 11:21
11
$\begingroup$

In a paper of Leep and Shapiro, the following theorem is proved without appealing to Hasse--Weil or Cauchy--Davenport/Vosper: Let $F$ be an arbitrary field and let $G$ be a subgroup of $F^{\ast}$ of index $3$. Then $G+G=F$ except when $|F| = 4, 7, 13$, or $16$. It seems that when $F$ is finite, the proof given there is due to Berrizbeitia.

The reference is
Multiplicative subgroups of index three in a field, Proc. Amer. Math. Soc. 105 (1989), 802--807.

$\endgroup$
3
$\begingroup$

The argument in this MO answer: Simple proofs for the existence of elliptic curves having a given number of points shows that the number of solutions to $y^2=f(x)$, $f$ a cubic, is at most $3p/2$. Taking a quadratic twist, it follows that the number of solutions is at least $p/2$. The equation $x^3+y^3=a$ can be put in Weierstrass form with a slightly messy but elementary calculation and the result follows.

$\endgroup$
2
$\begingroup$

Using character sums one can show that if $G$ is a subgroup of $\mathbb{F}_p^\times$ with $|G|>p^{\frac{k+1}{2k}}$, then every element of $\mathbb{F}_p^\times$ is a sum of $k$ elements from $G$. In particular, it follows for $p\geq 83$ that every element of $\mathbb{F}_p^\times$ is a sum of two cubes.

$\endgroup$

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.