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We are interested in estimating the size of a certain sumset in $\mathbb{Z}/p^2\mathbb{Z}$. Let $p$ be an odd prime, $g$ a primitive root modulo $p^2$, and $A=\langle g^p\rangle$ the unit subgroup of order $p-1$. Experimentation suggests strongly that if $p \geq 61$, then the size of the sumset $3A = A + A + A$ approximates the order of the ring. This observation is due to Felipe Voloch (see the original question). In particular, we conjecture that $$|3A|\geq p^2-1,$$ for such $p$. It is known that, as sets, $4A = \mathbb{Z}/p^2\mathbb{Z}$. Additionally, numerical evidence suggests that $3A \supseteq (\mathbb{Z}/p^2\mathbb{Z})\setminus\{0\}$, and that $p\equiv 1\pmod{3}$ implies strict containment (i.e., $3A = \mathbb{Z}/p^2\mathbb{Z}$). Perhaps these observations could provide a foothold toward the result. Is there work that addresses this phenomenon, or any existing progress toward an explanation of the estimate above?

(Edit: The remark about "strict containment" refers to the $\supseteq$ relation. That is, $3A \supsetneq (\mathbb{Z}/p^2\mathbb{Z})\setminus\{0\}$ whenever $p\equiv 1\pmod{3}$, which implies that $3A=\mathbb{Z}/p^2\mathbb{Z}$.)

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Is it known that if $p\equiv 1\pmod{3}$ then $0\in 3A$? I was unclear whether you knew this for a fact or just had strong numerical evidence. – Thomas Bloom Jul 5 '13 at 18:59
@Thomas Bloom If $p \equiv 1 \pmod 3$, then $1+a^p+(a^2)^p=0 \pmod {p^2}$ if $a$ has order $3$ $\pmod p$. – Felipe Voloch Jul 5 '13 at 19:12
Wouldn't "strict containment" be inequality, rather than equality? – Gerry Myerson Jul 5 '13 at 23:39
@Gerry Myerson The original remark was slightly ambiguous; I've edited the question to clarify. – Bob Lutz Jul 6 '13 at 8:07
@Felipe Voloch That is a very clean proof. It explains some of the numerical results we've been seeing. – Bob Lutz Jul 6 '13 at 8:09
up vote 2 down vote accepted

Using some results of Shkredov et al (following Heath-Brown and Konyagin), it's possible to show that $|A+A+A|\gg |A|^{11/6}/(\log|A|)^{1/4}$. I posted more details in the old thread. Shkredov's results are rather complicated, so it would be really nice to see a slick proof.

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Since I don't have enough rep to comment on the Bob's question, I'll ask a question here. Does your numerical data show that $|A+A|\approx p^2$ for large $p$? – Brendan Murphy Jul 11 '13 at 9:18
A cursory answer to your question: After testing primes up to $p=7499$, the ratio $|A+A|/p^2$ seems to hover around $0.39$, never deviating more than $0.015$ in either direction after $p=2531$ (and probably for a while before that as well). I am skipping 20 primes at a time, since the computation becomes sluggish rather quickly, but this behavior seems consistent. In particular, I have not seen a trend of increase or decrease in the ratio, but this might change, of course, as $p$ becomes truly large. – Bob Lutz Jul 12 '13 at 5:36

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