sum of quadratic residues We know that when $q$ is a prime $1(\text{mod } 4)$ then
$$
\sum_{R_i}R_i = q(q-1)/4,
$$
where $R_i$ are all the quadratic residues $\text{mod } q$.
I was curious if there is a similar (or some sort of) formula 
when $q$ is a prime $3(\text{mod } 4)$.
Thanks!
 A: Yes, there's a formula, though not quite as simple: the sum is
$$
\frac12 \left( {q \choose 2} - \frac2w qh \right),
$$
where $h$ is the class number of the quadratic imaginary field
${\bf Q}(\sqrt{-p})$, and $w$ is the number of roots of unity
in that field, so that the factor $2/w$ is just $1$ except for $q=3$
when $2/w = 1/3$.
This is obtained by writing the sum of the residues as 
$\frac12 \sum_{r=1}^{p-1} (1+(r/p)) r$, where $(r/p)$ is the Legendre symbol.
The $\sum_r r$ part of this gives $q \choose 2$, and the formula $-2qh/w$ for the
$\sum_r (r/p) r$ part is classical (the factor $2/w$ arises via the
Dirichlet class number formula, see for instance the $d<0$ part of
equation (2) in MathWorld's
"Class Number" page).
[The formula for $q \equiv 1 \bmod 4$
is simple because in that case the identity $(r/p) = (-r/p) = ((p-r)/p)$ yields
$\sum_{r=1}^{p-1} (r/p) r = 0$ by symmetry.]
A: This is a comment too long to fit in the usual format. For any $k$, denote by $r_k$ the remainder of the division of $k^2$ by $q$. Then , the sum we are considering is $S=\displaystyle\sum_{k=1}^{2b+1} r_k $, where $b$ is defined by $q=4b+3$. Leaving aside the case $q=3$, we see that $b$ cannot be divisible by $3$.
Note that modulo $q$, one has
$$
S =\sum_{k=1}^{2b+1} r_k \equiv \sum_{k=1}^{2b+1} k^2
=\frac{(2b+1)(2b+2)(4b+3)}{6}=q \frac{(b+1)(2b+1)}{3} \equiv 0 \ ({\sf mod}\ q).
$$
So $S$ is a multiple of $q$. If we put $T=\frac{S}{q}$, there seems to be no simple
formula for $T$.
