4. When $m$ and $n$ are both odd numbers, the equation
$$
x^{2}+4 m x-3 n x+4 m^{2}+n^{2}+9=0
$$

has two roots that are prime numbers. These two roots are ( ).
(A) 2 and 11
(B) 3 and 7
(C) 2 and 5
(D) 3 and 11
4. (A).

Let the two roots be $x_{1}, x_{2}$. Then $x_{1}+x_{2}=3 n-4 m$.
Since $m, n$ are odd numbers, $3 n-4 m$ is odd, and $x_{1}, x_{2}$ are prime numbers,
Therefore, one of $x_{1}, x_{2}$ must be the even number 2.
Substituting 2 and rearranging, we get
$4(m+1)^{2}+(n-3)^{2}=0$, solving this gives $m=-1, n=3$.
Substituting these values into the original equation and simplifying, we get
$x^{2}-13 x+22=0$, solving this gives $x_{1}=2, x_{2}=11$.