$-2<\alpha<1<\beta<2$

Let $\alpha = \frac{-p -\sqrt{p^2+4}}{2}, \beta = \frac{-p+\sqrt{p^2+4}}{2}$ be the two roots in question. If we require $-2 < \alpha < 1 < \beta < 2$ to hold, then we have:

$-2 < \frac{-p -\sqrt{p^2+4}}{2} < 1$ (i)

$1 < \frac{-p +\sqrt{p^2+4}}{2} < 2$ (ii)

If we solve for $p$ in (ii), then we obtain:

$p+2 < \sqrt{p^2+4} < p+4 \Rightarrow p^2 + 4p + 4 < p^2+4 < p^2+8p+16 \Rightarrow \boxed{-\frac{3}{2} < p < 0}$

of which $\boxed{p=-1}$ is the only integer solution.