WakeUp wrote: Let $a$ and $b$ be positive integers. Show that if $a^3+b^3$ is the square of an integer, then $a + b$ is not a product of two different prime numbers. Write $a+b=pq$ and $a^3+b^3=(a+b)^3-3ab(a+b)=u^2$ and so $pq(p^2q^2-3ab)=u^2$ So $p|3ab$ and $q|3ab$ and since we cant have $p=q=3$, we get that one of $p,q$ at least divides one of $a,b$, wlog say $p|a$ with $p>3$ So $a=kp$ and $b=p(q-k)$ with $q>k$ and we get $pq(p^2q^2-3kp^2(q-k))=u^2$ and so $pq(q^2-3k(q-k))=v^2$ and so $q|3k^2$ and so either $q|k$, which is impossible since $q>k$, either $q=3$ So $q=3$ So $a=kp$ and $b=p(3-k)$ with $k\in\{1,2\}$ and $p(3-k(3-k))=w^2$ : $k=1$ implies $p=w^2$, impossible $k=2$ implies $p=w^2$, impossible Hence the result.