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Jan 9 at 19:42 vote accept Bobby Morelli
Jan 4 at 19:10 comment added Sandeep Silwal beautiful answer!
Jan 4 at 17:22 history edited Emil Jeřábek CC BY-SA 4.0
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Jan 4 at 17:20 comment added Emil Jeřábek Your suggestion is on the right track, though. Whatever the maximal gap size is, I believe $p+1$ translates (non-consecutive) of Singer’s set do cover the whole cyclic group; I’ve updated the answer.
Jan 4 at 17:16 history edited Emil Jeřábek CC BY-SA 4.0
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Jan 4 at 14:33 comment added Sean Eberhard Good point, that argument doesn't quite work.
Jan 4 at 14:11 comment added Emil Jeřábek Ah, this is good to know. Though I don’t quite follow the argument. Why cannot there be a gap of size larger than about $\sqrt N$? I can see that if there is a gap of size $g$, then we get a Sidon set of size $p+1$ in $[0,p^2+p+1-g)$, which implies $p+1<\sqrt{p^2+p+1-g}+\sqrt[4]{p^2+p+1-g}+1$ by the Lindström bound. But this only bounds $g$ by $2p^{3/2}\sim2N^{3/4}$ or so, not $N^{1/2}$.
Jan 4 at 12:12 comment added Sean Eberhard One can improve this to $\sim \sqrt N$. Choose $p$ minimal so that $p^2+p+1 > N$ and use the Singer construction of a perfect difference in the cyclic group of order $p^2+p+1$. There cannot be a gap larger than about $N^{1/2}$, by the Erdős--Turan bound.
Jan 4 at 11:29 history answered Emil Jeřábek CC BY-SA 4.0