Actually, as you've written it it's not true, but with appropriate modifications it is: If p = 3 (mod 4) let S = sum_{k=1}^{p-1} k^2 mod p. Then S = sum_{k=1}^{p-1} k (1+(k/p)), where the inner (k/p) is the Legendre symbol. We count each square twice in S, which cancels the 2 in the denominator of (1+(k/p))/2 Using the class number formula that sum_{k=1}^{p-1} k (k/p) = -p*h(-p), we get S = p(p-1)/2- p*h(-p) Victor On Mon, Mar 15, 2010 at 7:54 PM, Victor Miller <victorsmiller@gmail.com> wrote:
David, Yes, it is true and well-known. It's more usually written in the form:
if p = 3 mod 4, then
h(-p) = -(1/p) sum_{m=1}^{p-1} m (m/p),
where (m/p) is the Legendre symbol (qudratic residue symbol). For example, see "Multiplicative Number Theory" by Davenport, page 53 (first edition).
Victor
On Mon, Mar 15, 2010 at 7:23 PM, David Wilson <davidwwilson@comcast.net> wrote:
Let f(n) = SUM(0 <= k < n; (k^2) mod n)
where mod is the remainder function.
It looks as if, for prime p >= 5
f(p) =
1 if p == 1 (mod 4) 2c+1 if p == 3 (mod 4)
where c is the class number of Q(sqrt(-p))
Is this true? Well known?
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