If $a,b$ are natural numbers such that $a^3+b^3$ is a perfect square, prove that $a+b$ can't be equal to the product of two different prime numbers.
I've tried to find some congruencies that would help me prove a contradiction, but that came to nothing.
I think the solution has something to do with the fact that $a^3+b^3=(a+b)(a^2-ab+b^2)$
But, I'm not sure where to go from there, could someone tell me if I'm on the right path and give me a clue, thanks.
We have $$a^3+b^3=(a+b)((a+b)^2-3ab)$$ If $p$ is one of the two primes that divide $a+b$ then since $a^3+b^3$ is a square, $$p|((a+b)^2-3ab)$$ so $p|3ab$ and since there are two primes we can assume that $p\neq 3$. Then say $p|a$ as $p|a+b$ we have $p|b$ also. Let $a=px$ $b=py$ then
$$x^3+y^3=(x+y)((x+y)^2-3xy)$$ which is divisible by $q^2$, $q$ being the other prime dividing $a+b$ and is also divisible at least once by $p$. Now that $x+y=q$ so we have
$$x^3+y^3=q(q^2-3xy)$$
and thus $q|3xy$ if $q|x$ or $|y$ then it divides both. And $x+y$ is not prime. Thus $q=3$.
So finally $x+y=3$, and therefore $x=2, y=1$. ($x=0$ is clearly impossible.)
and we have
$$2^3+1=3(9-6)$$ however this is not divisible by the other prime $p$.