Answer
$yes\,\,,\left \{ A_{0},A_{1},A_{2},A_{3} \right \}\,is\,a\,partition\,of\,\,\mathbb{Z} $
$A_{0}=\left \{ n\in \mathbb{Z}\mid n=4k,\,\,for\,some\,integer\,k \right \}$
$A_{1}=\left \{ n\in \mathbb{Z}\mid n=4k+1,\,\,for\,some\,integer\,k \right \} $
$A_{2}=\left \{ n\in \mathbb{Z}\mid n=4k+2,\,\,for\,some\,integer\,k \right \}$
$A_{3}=\left \{ n\in \mathbb{Z}\mid n=4k+3,\,\,for\,some\,integer\,k \right \}$
$\mathbb{Z}\,\,the\,\,set\,of\,all\,integers\,$
Work Step by Step
$according\,\,to\,\,the\,quotient-remainder\,\,theorem$
$any\,integer\,\,n\,\,can\,\,be\,represented\,\,in\,exactly\,\,one\,of\,the\,\,four\,forms\,$
$A_{0}=\left \{ n\in \mathbb{Z}\mid n=4k,\,\,for\,some\,integer\,k \right \}$
$A_{1}=\left \{ n\in \mathbb{Z}\mid n=4k+1,\,\,for\,some\,integer\,k \right \} $
$A_{2}=\left \{ n\in \mathbb{Z}\mid n=4k+2,\,\,for\,some\,integer\,k \right \}$
$A_{3}=\left \{ n\in \mathbb{Z}\mid n=4k+3,\,\,for\,some\,integer\,k \right \}$
$and\,\,this\,implies\,\,that\,\,$
$A_{0}\cup A_{1}\cup A_{2}\cup A_{3}=\mathbb{Z}$
$and\,\,A_{0},A_{1},A_{2},A_{3}\,\,are\,mutually\,\,dis\! joint\,\,(as\,any\,integer\,can\,not\,be\,in\,any\,\,two\,\,of\,the\,sets)$
$so\,\,,\left \{ A_{0},A_{1},A_{2},A_{3} \right \}\,is\,a\,partition\,of\,\,\mathbb{Z} $