Vì \(x,y\in N\Rightarrow2y+1\) lẻ
Do đó \(\left(2y+1\right)\left(x-4\right)=2\cdot5=1\cdot10\)
\(\left\{{}\begin{matrix}2y+1=5\\x-4=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=6\\y=3\end{matrix}\right.\rightarrow\left(6;3\right)\\ \left\{{}\begin{matrix}2y+1=1\\x-4=10\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=14\\y=0\end{matrix}\right.\rightarrow\left(14;0\right)\)
Vậy PT có nghiệm \(\left(x;y\right)\) là \(\left(6;3\right);\left(14;0\right)\)