Difference of primes

Rearrange the equation as $3q=p^{2}-64=p^{2}-8^{2}=(p+8)(p-8)$ .

Since $q$ is prime, $3q$ has only two factors $3$ and $q$ , so either $p+8=3$ or $p-8=3$ .

The first equation tells us that $p=-5$ which cannot occur as $p>0$ , so we conclude that $p$ satisfes the second equation and is equal to $11$ . Substituting $p=11$ into $3q=(p+8)(p-8)$ gives $3q=19*3$ and so $q=19$ . Hence $p+q=11+19=\boxed {30}$ .

Logically....>given that p and q are prime no.and according to eqn p^2 must be greater tham 3q hence only 11 is a no whose square will be greater hence p=11.now. 11^2=121 121-64=57 57/3=19 Hence q=19

$\Large 11^{2}-(3 \times 19) = 64$

Since both are positive p must be greater than 7 .... Subbed first possible value and low and behold it worked.