Ta có \(x^3-y^3-2y^2-3y-1=0\Leftrightarrow x^3=y^3+2y^2+3y+1\)
Ta có \(\left(y-1\right)^2=y^3-3y^2+3y-1=y^3+2y^2+3y+1-5y^2-2=y^3+2y^2+3y+1-\left(5y^2+2\right)< y^3+2y^2+3y+1\)(vì 5y2+2>0)\(\Leftrightarrow\left(y-1\right)^3< x^3\)(1)
Tương tự \(\left(y+1\right)^3=y^3+3y^2+3y+1=y^3+2y^2+3y+1+y^2\ge y^3+2y^2+3y+1\)(vì \(y^2\ge0\))\(\Leftrightarrow x^3\le\left(y+1\right)^3\)(2)
Từ (1),(2)\(\Rightarrow\left(y-1\right)^3< x^3\le\left(y+1\right)^3\Leftrightarrow y-1< x\le y+1\)
Vì x,y nguyên
Suy ra \(\left[{}\begin{matrix}x=y\\x=y+1\end{matrix}\right.\)
_Nếu x=y
\(x^3-y^3-2y^2-3y-1=0\Leftrightarrow y^3-y^3-2y^2-3y-1=0\Leftrightarrow2y^3+3y+1=0\Leftrightarrow2y^2+2y+y+1=0\Leftrightarrow2y\left(y+1\right)+\left(y+1\right)=0\Leftrightarrow\left(y+1\right)\left(2y+1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}y+1=0\\2y+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=-1\left(tm\right)\\y=-\dfrac{1}{2}\left(ktm\right)\end{matrix}\right.\)\(\Leftrightarrow y=-1\Leftrightarrow x=y=-1\)
_Nếu x=y+1
\(x^3-y^3-2y^2-3y-1=0\Leftrightarrow\left(y+1\right)^3-y^3-2y^2-3y-1=0\Leftrightarrow y^3+3y^2+3y+1-y^3-2y^2-3y-1=0\Leftrightarrow y^2=0\Leftrightarrow y=0\Leftrightarrow x=1\)
Vậy (x;y)=(-1;-1);(1;0)
