If $p$ is a prime and both roots of $x^{2} +px−444p=0$ are integers what is $p$ 
If p is a prime and both roots of $x^{2}+px−444p=0$ are integers, what
  is $p$

I got that for the roots to be integers the discriminant must be a perfect square. Thus, $p(p + 1776)$ must be a perfect square.
However, at this point I am stuck and do not know how to proceed other than trying values for $p$ and checking what works.
 A: HINT:
We have $\displaystyle q^2=p(p+1776)$ where $q$ is some integer
As $p$ is prime, $\displaystyle  p$ must divde $q,q=pr$(say)
$\displaystyle\implies 1776p=p^2(r^2-1)\iff \frac{1776}p=r^2-1$ which is an integer 
Check for the prime factors of $1776$
A: Factor it as $(x-a)(x-b)$. Then, $ab=-444p$, so one of $a,b$ is a multiple of $p$. But since $a+b =-p$, they are both multiples of $p$.So $p$ is a divisor of $444$(because $-444p$ would be a multiple of $p^2$)
A: $ x^2\!+px = 444p\,\Rightarrow \,p\mid x^2\,\Rightarrow\,p\mid x\,\Rightarrow\ p^2\mid 444p\,\Rightarrow\, p\mid 444\,\Rightarrow\, X^2\! + X = 444/p,\ X = x/p,\, $ which has roots $\,m,n\,$ with $\ \color{#C00}{m+n} = -1,\,\ mn = -444/p = \color{#c00}{- 4\cdot 3}\cdot 37/p,\ $ so $\ \color{#c00}{m,n},p \, =\, \ldots$
A: By Eisenstein's criterion, the polynomial $x^2 + px - 444p$ is irreducible unless $p^2|444p$.
If both roots of the polynomial are integers, it is not irreducible and hence
$$p^2|444p \;\;\iff\;\; p | 444 \;\;\implies\;\; p = 2, 3 \text{ or } 37$$
So we have only 3 primes to check.
