A number theory problem by Áron Bán-Szabó

Since the LHS is positive, it follows that $(p-q)^3 > 0$ or equivalently $p>q$ .

The condition clearly yields that $p-q | p+q$ from which $p-q | p+q+p-q$ or $p-q | 2p$ . For $p-q>0$ we have 3 cases:

Case 1.: $p-q=1$

This would imply the LHS to be 1 but the sum of two primes cannot be 1, so we have reached a contradiction.

Case 2.: $p-q=2$

In this case $p+q=2^3=8$ so we have the following system of equations:

$p-q=2$

$p+q=8$

This system of equations has exactly one solution, namely $p=5$ and $q=3$ .

Case 3.: $p-q=p$ or $p-q=2p$

In both of the possibilities we would obtain $p-q \geq p$ which is impossible for $q>0$ .

Hence the only solution is $p=5$ and $q=3$ so the answer is $5+3=8$ .

Let p > q and p-q = 2n if both p and q are odd. Then (p - q)^3 = 8n^3. p +q = p - q + 2q = 2n + 2q. So 8n^3 = 2n + 2q, and dividing by 2, 4n^3 = n + q. Hence n|q, but q is prime, so that n = 1 or n = q. If n = q, p = 2n + q = 3n, which cannot be, since p is prime, and not divisible by 3. Therefore n = 1,, p -q = 2, p + q =8. Ed Gray