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Aug 21, 2021 at 10:51 comment added mathworker21 thanks. By the way, the reference in the other answer gets to $3/7$ in (essentially) the same way and then pushes past that by using bounds for exponential sums over multiplicative subgroups.
Aug 21, 2021 at 10:38 comment added Random @mathworker21 Oops, you're right. Fixed that
Aug 21, 2021 at 10:37 history edited Random CC BY-SA 4.0
3/8 to 3/7
Aug 21, 2021 at 9:29 comment added mathworker21 $+1$. "As stated in the post, it is sufficient to show $|A_d| \le p^{3/8+\epsilon}$". Do you mean $p^{3/7+\epsilon}$?
Aug 21, 2021 at 1:27 history answered Random CC BY-SA 4.0