smc

For integer roots, we need the discriminant, which is $p^2-4\cdot 1\cdot (-444p)=p^2+1776p=p(p+1776)$, to be a perfect square. Now, this means that $p$ must divide $p+1776$, as if it did not, there would be a lone prime factor of $p$, and so this expression could not possibly be a perfect square. Thus $p$ divides $p+1776$, which implies $p$ divides $1776=2^4\cdot 3\cdot 37$, so we must have $p=2$, $3$, or $37$. It is easy to verify that neither $p=2$ nor $p=3$ make $p(p+1776)$ a perfect square, but $p=37$ does, so the answer is $31<p\le 41$, which is $\boxed{31<p\le 41}$.