I'm trying to find all positive integers $a,b$ and $c$ that satisfy these conditions: \begin{cases} a^3 - b^3 - c^3 = 3abc \\ a^2 = 2(b+c) \end{cases}
When I saw that I remember this algebraic alliance: $$(a-b-c)(a^2+b^2+c^2+ab+ac-bc)=a^3-b^3-c^3-3abc$$ Thus $a=b+c$ or $a^2+b^2+c^2+a(b+c)=bc$ I couldn't do anything more. Is there any other answer? This question was asked 32 years ago at Iranian Math Competition.
$a-b-c=0$ implies $a=b+c$ (and not $a=b=c$). Substitute into the second equation to deduce $a^2=2a$, which yields $a=2$, since we rule out $a=0$. The only positive integer solutions to $2=b+c$ are $b=1$, $c=1$, so one solution to the system is $(a,b,c)=(2,1,1)$.
The second factor $a^2+b^2+c^2 +ab+ac-bc$ is strictly positive for $a,b,c$ positive, since it exceeds $b^2-bc+c^2 > b^2-2bc+c^2=(b-c)^2\ge0$. Therefore there are no other solutions.