Please help to prove the following.

Question

a,b and c are integers and we know that a+b+c=(a-b)(b-c)(c-a)

Prove, that a+b+c is divisible by 27.

Thank you very much.

Answer 1

Think of it modulo $3$.

It is obvious that $a + b + c \equiv 0 \pmod 3$.

So, since the variables are symmetrical, we have as only possibilities

$$(a, b, c) \equiv (0,0,0) \ \ (1, 1, 1) \ \ (0, 1, 2) \ \ (2, 2, 2) \pmod 3$$ On the other hand the tuple $(0, 1, 2)$ must be eliminated because otherwise the right hand side would not be divisible by $3$

Finally it is trivial to check that for each of the $3$ tuples remaining, the right hand side is the product of $3$ numbers, all divisible by $3$ and this implies that $a + b + c$ is divisible by $27$