, 2016

We show that all of $a, b$ and $c$ have the number $3$ as a common factor. First suppose that one of $a, b, c$ is divisible by $3$. By symmetry, we may assume that $a \equiv 0 \pmod{3}$. Then the equation implies $bc \equiv (b-c)(b+c)(c-b) \pmod{3}$. If neither of $b$ and $c$ is divisible by $3$, this gives $bc \equiv 0 \pmod{3}$, which is a contradiction. On the other hand, if either of $b$ and $c$ is divisible by $3$, then from $bc \equiv (b-c)(b+c)(c-b) \pmod{3}$ we see that they both are divisible by $3$, which means that $a, b$ and $c$ are all divisible by $3$.

Now we are left with the case that $a, b, c$ are each $\pm 1 \pmod{3}$. If $a \equiv b \equiv c \pmod{3}$, then clearly the left-hand side of the equation is divisible by $3$, while the right-hand side is not, so we have a contradiction. In the opposite case that two of $a, b$ and $c$ are equal and the third one is distinct modulo $3$, clearly the left-hand side of the equation is not divisible by $3$, while the right-hand side is divisible. We conclude that $a, b$ and $c$ are all divisible by $3$.