[math-fun] a bracelet sum

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 Wouter.