3. Given that $p$ is a prime number, and the equation
$$
x^{2}+p x-444 p=0
$$

has two integer roots. Then $p=$ $\qquad$
3.37.

According to
$$
x^{2}=p(444-x)
$$

we know that $p(444-x)$ is a perfect square.
Since $p$ is a prime number, it follows that $p\left|x^{2} \Rightarrow p\right| x$.
Let $x=n p(n \in \mathbf{Z})$ and substitute into equation (1) to get
$$
(n p)^{2}=p(444-n p) .
$$

Since $p \neq 0$, we have $n^{2} p=444-n p$, which is
$$
n(n+1) p=2^{2} \times 3 \times 37 \text {. }
$$

Therefore, $p=37$.