This is a Gauss sum. As I recall, it’s not too hard to square the expression and show that one gets p in the first case and -p in the second. This means that the first expression is plus-or-minus Sqrt[p] in the first case and plus-or-minus I Sqrt[p] in the second case. The hard part is getting rid of the sign-ambiguity. Jim On Sat, Jun 8, 2019 at 10:18 AM Wouter Meeussen <wouter.meeussen@telenet.be> wrote:
is it easy (or trivial) to prove that
Sum[Exp[2 \[Pi] I Mod[(6k+1)^2, p]/p], {k,1,p}]
equals Sqrt[p] for p prime and Mod[p,4]==1 and I*Sqrt[p] for p prime and Mod[p,4]==3 ?
It also has a combinatorial significance cfr. https://math.stackexchange.com/questions/2899878
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