How about S? S has (p-1)/2 elements in it. To find how many are greater than p/2, you can instead find how many are less than p/2.Oxymoron said:p exponent
3 1
5 1
7 2
11 3
13 3
17 4
19 5
Yes. Now can you write the floor of |S|/2 in a nother way? You know |S|=(p-1)/2.Oxymoron said:The number of elements in S, denoted |S| will always be equal to (p-1)/2 as you said, and the number of these elements less than p/2, denoted by #<p/2 is always equal to the floor of |S|/2.
I don't know what you mean by "negative least residue". In the Gauss Lemma, you take the least positive residues of the elements of S, but when p=2 they are already reduced.Oxymoron said:So this becomes a problem about showing that the number of elements in the set of negative least residues less than p.