Inspired by Satyajit Mohanty

Notice that if $x \equiv y \mod p$ for least positive $y$ , then $\left\lfloor \frac{x}{p} \right\rfloor=\frac{x}{p}-\frac{y}{p}$ . Furthermore, we will be using the Legendre Symbol, denoted $\left(\frac{k}{p}\right)$ .

Now choose least positive $a_n$ and $b_n$ such that $3n^2 \equiv a_n \mod p$ and $n^2 \equiv b_n \mod p$ for all $n$ , $1 \leq n \leq p-1$ . Then we have $\displaystyle S(p)=\sum_{n=1}^{p-1} \left( \left \lfloor \frac{3n^2}{p} \right \rfloor - 3 \left \lfloor \frac{n^2}{p} \right \rfloor \right)=$

$\displaystyle \sum_{n=1}^{p-1} \left( \frac{3n^2}{p} - \frac{a_n}{p} -\frac{3n^2}{p} + \frac{3b_n}{p}\right)=$

$\displaystyle \sum_{n=1}^{p-1} \left( \frac{3b_n}{p} - \frac{a_n}{p}\right)=$

$\displaystyle \frac{3}{p}\sum_{n=1}^{p-1} b_n - \frac{1}{p}\sum_{n=1}^{p-1} a_n$

Here notice that $b_n$ cycles through the quadratic residues modulo $p$ twice (since $b_n=b_{p-n}$ ). Furthermore, since $p=101$ and by Quadratic Reciprocity $\left(\frac{3}{101}\right)=\left(\frac{101}{3}\right)=\left(\frac{2}{3}\right)=-1$ , we know $a_n$ must similarly cycle through the quadratic nonresidues twice. Therefore, our sum becomes

$\displaystyle \frac{6}{p}\sum_{\left(\frac{b}{p}\right)=1} b - \frac{2}{p}\sum_{\left(\frac{a}{p}\right)=-1} a$

Due to the known result that $\sum_{\left(\frac{b}{p}\right)=1} b=\frac{p(p-1)}{4}$ when $p \equiv 1 \mod 4$ , our sum becomes

$\displaystyle \frac{6}{p}\left(\frac{p(p-1)}{4}\right) - \frac{2}{p}\left(\frac{p(p-1)}{4}\right)=$

$\displaystyle \frac{3(p-1)}{2} - \frac{(p-1)}{2}=p-1$

Since $p=101$ , we have $S(101)=\boxed{100}$

Proof that

$\displaystyle \sum_{\left(\frac{b}{p}\right)=1} b=\frac{p(p-1)}{4}$ when $p \equiv 1 \mod 4$

Proof: Since $p \equiv 1 \mod 4$ , we know $\left(\frac{-1}{p}\right)=1$ . Thus, for each quadratic residue $b$ , we know $p-b$ is another distinct quadratic residue. Since there are $\frac{p-1}{2}$ such $b$ , there are $\frac{p-1}{4}$ pairs $(b, p-b)$ , each of which sum to $p$ . Thus, grouping these pairs together, we have

$\displaystyle \sum_{\left(\frac{b}{p}\right)=1} b= \sum_{(b, p-b)} p=\frac{p(p-1)}{4}$

Corollary $\displaystyle \sum_{\left(\frac{a}{p}\right)=-1} a=\frac{p(p-1)}{4}$

Since $\displaystyle \sum_{c=1}^{p-1} c= \frac{p(p-1)}{2}$ , we have

$\displaystyle \sum_{\left(\frac{a}{p}\right)=-1} a=\sum_{c=1}^{p-1} c - \sum_{\left(\frac{b}{p}\right)=1} b= \frac{p(p-1)}{2} - \frac{p(p-1)}{4}=\frac{p(p-1)}{4}$