Interesting Sum Congruence Let $5\mid a$, $\gcd(a,b)=1$, and $b\equiv 2\bmod 5$. How can one show that $\sum_{k=1}^{a}k\lfloor\frac{kb}{a}\rfloor\equiv 2\bmod 5$?
Similarly, can we show that if instead $b\equiv 3\bmod 5$, then $\sum_{k=1}^{a}k\lfloor\frac{kb}{a}\rfloor\equiv 3\bmod 5$?
 A: Since $(a,b)=1$, $\left\{\frac{kb}{a}\right\}$ ranges over $\left\{\frac1a,\frac2a,\dots,\frac{a-1}{a}\right\}$. Therefore,
$$
\begin{align}
\sum_{k=1}^{a-1}\left\{\frac{kb}{a}\right\}^2
&=\sum_{k=1}^{a-1}\left\{\frac{k}{a}\right\}^2\\
&=\frac{(2a-1)(a-1)}{6a}\tag{1}
\end{align}
$$
However, we also have $\left\{\frac{kb}{a}\right\}=\frac{kb}{a}-\left\lfloor\frac{kb}{a}\right\rfloor$. Therefore,
$$
\begin{align}
\sum_{k=1}^{a-1}\left\{\frac{kb}{a}\right\}^2
&=\sum_{k=1}^{a-1}\left(\frac{kb}{a}-\left\lfloor\frac{kb}{a}\right\rfloor\right)^2\\
&=\frac{b^2}{a^2}\sum_{k=1}^{a-1}k^2-2\frac{b}{a}\sum_{k=1}^{a-1}k\left\lfloor\frac{kb}{a}\right\rfloor+\sum_{k=1}^{a-1}\left\lfloor\frac{kb}{a}\right\rfloor^2\\
&=\frac{b^2}{a^2}\frac{(2a-1)(a-1)a}{6}-2\frac{b}{a}\sum_{k=1}^{a-1}k\left\lfloor\frac{kb}{a}\right\rfloor+\sum_{k=1}^{a-1}\left\lfloor\frac{kb}{a}\right\rfloor^2\tag{2}
\end{align}
$$
Subtracting $(2)$ from $(1)$ and multiplying by $a$, we get
$$
\begin{align}
2b\sum_{k=1}^{a-1}k\left\lfloor\frac{kb}{a}\right\rfloor
=\frac{(b^2-1)(2a-1)(a-1)}{6}+a\sum_{k=1}^{a-1}\left\lfloor\frac{kb}{a}\right\rfloor^2\tag{3}
\end{align}
$$
If $m\mid a$ and $(m,2b)=1$, then $(3)$ implies
$$
\sum_{k=1}^{a-1}k\left\lfloor\frac{kb}{a}\right\rfloor
\equiv\frac{b^2-1}{12b}\pmod{m}\tag{4}
$$
For $m=5$ and $b\equiv2\pmod{5}$, we get $\frac{b^2-1}{12b}\equiv2\pmod{5}$
For $m=5$ and $b\equiv3\pmod{5}$, we get $\frac{b^2-1}{12b}\equiv3\pmod{5}$

Clarification
Since $(m,2b)=1$, Bezout's Identity says that there are $x,y\in\mathbb{Z}$ so that
$$
mx+2by=1\tag{5}
$$
Squaring $(5)$ gives
$$
m(mx^2+4xby)+4b(by^2)=1\tag{6}
$$
Equation $(6)$ implies
$$
4b(by^2)\equiv1\pmod{m}\tag{7}
$$
$(7)$ says that $4b$ has an inverse mod $m$.
If $3\not\mid m$, then $(m,3)=1$ and $3$ has an inverse mod $m$.
If $3\mid m$, then $3\not\mid b$. Therefore, $3$ divides either $b-1$ or $b+1$, and thus $3\mid b^2-1$.
Therefore, whether or not $3\mid m$, $(4)$ makes sense.
