Lời giải:
\(\begin{array}{l}DK:x,y\in\mathbb{Z}\\y=\sqrt{x^2+2x+4}=\sqrt{(x+1)^2+3}\ge\sqrt{3}\\\to y\ge\sqrt{3}\\Ma:y\in\mathbb{Z}\\\to y\ge2\\pt\leftrightarrow y^2=x^2+2x+4\\\leftrightarrow y^2=(x+1)^2+3\\\leftrightarrow (x+1-y)(x+y+1)=-3\\\to x-y+1,x+y+1\in U(-3)=\{\pm1;\pm3\}\\* \begin{cases}x-y+1=1\\x+y+1=-3\\\end{cases}\\\to \begin{cases}x=-2\\y=-2\\\end{cases}(KTM)\\*\begin{cases}x-y+1=-1\\x+y+1=3\\\end{cases}\\\to \begin{cases}y=2\\x=0\\\end{cases}(TMDK)\\*\begin{cases}x-y+1=3\\x+y+1=-1\\\end{cases}\\\to \begin{cases}y=-2\\x=0\\\end{cases}(KTM)\\*\begin{cases}x-y+1=-3\\x+y+1=1\\\end{cases}\\\to \begin{cases}x=-2\\y=2\\\end{cases}(TMDK)\\\to (x,y)=(-2;2),(0;2)\end{array}\)
