Number theory question: Prove $27\mid a+b+c$ if $(a-b)(b-c)(c-a)=a+b+c$.

Question

Integers $a,b$ and $c$ satisfy $(a-b)(b-c)(c-a)=a+b+c$. Prove $27\mid a+b+c$.

Answer 1

First, assume that $3 \nmid (a+b+c)$. Then, $3 \nmid (a-b)(b-c)(c-a)$, which means that $a,b,c$ must be distinct modulo $3$. However, we still get $a+b+c \equiv 0+1+2 \equiv 0 \pmod{3}$ which is a contradiction. Thus, $3 \mid (a+b+c)$, and thus $3 \mid (a-b)(b-c)(c-a)$.

WLOG let $3 \mid (a-b)$. It then follows that: $$3 \mid (a+b+c) \implies 3 \mid (a-b)+(2b+c) \implies 3 \mid (2b+c) \implies b \equiv c \pmod{3}$$

Thus, we get $a \equiv b \equiv c \pmod{3}$. This means that each of the factors $(a-b),(b-c),(c-a)$ is divisible by $3$. Hence: $$27 \mid (a-b)(b-c)(c-a) \implies 27 \mid (a+b+c)$$ Hence, proved.