Let a, b, c be real numbers, not all of them equal. Prove that if $$a+b+c=0$$, then $$a^2+ab+b^2 = b^2+bc+c^2 = c^2+ca+a^2$$. Prove the converse: if $$a^2+ab+b^2 = b^2+bc+c^2 = c^2+ca+a^2$$, then $$a+b+c=0$$.

For the forward direction, substituting $$a=-b-c$$ into $$a^2+ab+b^2$$ and simplifying gives $$b^2+c^2+bc$$, which by symmetry equals both $$b^2+bc+c^2$$ and $$c^2+ca+a^2$$, proving the three expressions are equal. For the converse, the equation $$a^2+ab+b^2 = b^2+bc+c^2$$ simplifies to $$a^2-c^2+b(a-c)=0$$, which factors as $$(a-c)(a+b+c)=0$$; since a, b, c are not all equal, $$a \ne c$$ is possible in general, so the only way this holds is $$a+b+c=0$$.