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"