Có bao nhiêu cặp số nguyên dương $\left( x;y \right)$ thỏa mãn...

ĐK: $\left\{ \begin{aligned}
& x>3y \\
& x;y\in {{\mathbb{N}}^{*}} \\
\end{aligned} \right.$.
Ta có $\ln \left( x-3y \right)+{{x}^{2}}+3{{y}^{2}}+y=x\left( 4y+1 \right)$
$\Leftrightarrow \ln \left( x-3y \right)=-{{x}^{2}}-3{{y}^{2}}-y+x\left( 4y+1 \right)$
$\Leftrightarrow \ln \left( x-3y \right)=-{{x}^{2}}-3{{y}^{2}}-y+3xy+xy+x$
$\Leftrightarrow \ln \left( x-3y \right)=\left( xy-3{{y}^{2}}-y \right)+\left( -{{x}^{2}}+3xy+x \right)$
$\Leftrightarrow \ln \left( x-3y \right)=y\left( x-3y-1 \right)-x\left( x-3y-1 \right)$
$\Leftrightarrow \ln \left( x-3y \right)=\left( y-x \right)\left( x-3y-1 \right)$
$\Leftrightarrow x-3y={{e}^{\left( y-x \right)\left( x-3y-1 \right)}}$ $\left( * \right)$.
Vì $x,y\in {{\mathbb{Z}}^{+}}$ nên $\left\{ \begin{aligned}
& x-3y>0\ge 1 \\
& y-x<0 \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& x-3y-1\ge 0 \\
& y-x<0 \\
\end{aligned} \right. $ $ \Rightarrow {{e}^{\left( y-x \right)\left( x-3y-1 \right)}}\le 1$.
Từ đó suy ra $\left\{ \begin{aligned}
& VT\ge 1 \\
& VP\le 1 \\
\end{aligned} \right.$.
Vậy21q| $\left( * \right)\Leftrightarrow VT=VP=1$ $\Leftrightarrow x-3y=1\Leftrightarrow x=3y+1$.
Vì $0<x<2023$ nên $0<3y+1<2023\Leftrightarrow \dfrac{-1}{3}<y<674$, mà $y\in {{\mathbb{Z}}^{+}}$ nên $y\in \left\{ 1;2;3;4;...;673 \right\}$.
Vậy có 673 cặp số cần tìm.