Suppose $ p(x)=2 $ has at least 2 distinct integer roots $ y_1,y_2 $. Then $ x $ is a integer root of equation $ p(x)=1 $, $ z $ is a integer root of equation $ p(x)=3 $, $ x\neq z $. It's easy to see that $ y_1-x, y_2-x, y_1-z,y_2-z | 1 $ And there are just 2 cases, and both also reach a contradiction!!. So Answer is "No"