show that if $\frac{2(m^2+n^2+n)+1}{m(2n+1)}=k$, k is the multiple of 3 $$\frac{2(m^2+n^2+n)+1}{(2n+1)m}=k$$
where k, m, and n are all integer.
show that $3|k$

Write it as
$$
2(m^2+n^2+n)+1=k(m(2n+1))
$$
Compute both sides mod $3$, that is, for $m,n=0,1,2$.
It turns out that there are two cases:


*

*The LHS is not zero and the RHS is zero. This cannot happen.

*The LHS is zero and the RHS is not zero. This implies that $k \equiv 0 \bmod 3$.
$$
\begin{array}{cccc}
m & n & 2(m^2+n^2+n)+1 & m(2n+1) \\
0 & 0 & 1 & 0 \\
0 & 1 & 2 & 0 \\
0 & 2 & 1 & 0 \\
1 & 0 & 0 & 1 \\
1 & 1 & 1 & 0 \\
1 & 2 & 0 & 2 \\
2 & 0 & 0 & 2 \\
2 & 1 & 1 & 0 \\
2 & 2 & 0 & 1 \\
\end{array}
$$
