Quadratic Primes

Let $a$ and $b$ be the roots of the equation.

$\Rightarrow a + b = p$ and $ab =q$

Since $q$ is prime, either of $a$ and $b$ must be $1$ and the other equal to $q$

Assume that $a = 1$ and $b = q$

$\Rightarrow q + 1 = p$

From this equation its clear that either of $p$ or $q$ must be even and the other odd , but $p$ and $q$ are primes , so

$\Rightarrow p =3$ and $q=2$

$\Rightarrow (p^3 + q^3)(p-q+1) = 70$

It is given that the roots are positive and integral.

(x^{2} + q = px

Here, x has to be positive and integral. We observe that the square of x + (a prime number) is divisible by x to satisfy the equation.

That is only possible when x is 1 or equal to q for positive integral value.

This gives us that q+1=p or q^2 + q=p

When x=1, p=3 and q=2.

So answer for this question is 70.