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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

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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 $u=v=0$.

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    $\begingroup$ 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 '14 at 11:12
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    $\begingroup$ There should be a simpler proof using the Cauchy Davenport inequality. (We might need Vosper's result which also characterizes when equality can happen) $\endgroup$ – Gjergji Zaimi Oct 24 '14 at 11:14
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    $\begingroup$ @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 '14 at 11:21
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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.

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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.

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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.

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