Discriminate

Assume that the first root is $x_1$ and the second is $x_2$

By Vieta Formula, we know that $x_1 + x_2 = p$ , $x_1 x_2 = q$

Remember that a prime number is a number that have factor $1$ and the number itself.

Hence, $q = x_1$ and $1 = x_2$

Since $q + 1 = p$ , $q$ and $p$ must be $3$ and $2$ ( The only prime number that satisfy the equation)

Finally, $3+2 = \color{#D61F06}{\boxed{5}}$

If the above quadratic equation is to have two distinct integer roots, then they satisfy:

$x = \frac{p \pm \sqrt(p^2 - 4q)}{2}$

which requires the discriminant to be a positive perfect square, or $p^2 - 4q = k^2$ ( $k \in \mathbb{N})$ , which factors as $(p+k)(p-k) = 4q$ . The number $4q$ has the prime factorization $2^2q^1$ that yields three positive integer divisor pairs: $(1,4q); (2,2q); (4,q)$ . We now examine these pairs under the following systems of equations:

$p + k = 4q; p - k = 1 \Rightarrow 2p = 4q + 1$ has no pair $(p,q)$ since an even integer (LHS) cannot equal an odd integer (RHS);

$p + k = 2q; p - k = 2 \Rightarrow 2p = 2q + 2 \Rightarrow p = q + 1$ , which is only satisfied for p = 3, q = 2.

$p + k = q; p - k = 4 \Rightarrow 2p = q + 4$ , which again is only satisfied for p = 3, q = 2 since the LHS is an even integer that forces the RHS to be even as well.

Hence $(p,q) = (3,2)$ is the only prime pair that works above with $p + q = 3 + 2 = \boxed{5}$ .