Junior Balkan Mathematical Olympiad

Let $a$, $b$, $c$ be integers and $n$ be a positive integer such that $$(a-b)(b-c)(c-a)+4=2\cdot 2016^n.$$ We set $a-b=-x$, $b-c=-y$ and we rewrite the equation as $$xy(x+y)+4=2\cdot 2016^n.\text{\hspace{1em}(1)}$$ If $n>0$, then the right hand side is divisible by $7$, so we have that $$xy(x+y)+4\equiv 0(\mathrm{mod}7)\text{\hspace{1em}(2)}$$ or $$3xy(x+y)\equiv 2(\mathrm{mod}7)\text{\hspace{1em}(3)}$$ or $$(x+y{)}^3-x^3-y^3\equiv 2(\mathrm{mod}7).\text{\hspace{1em}(4)}$$ Note that, by Fermat's Little Theorem, for any integer $k$ the cubic residues are $k^3\equiv -1,0,1(\mathrm{mod}7)$. It follows that in (1) some of $(x+y{)}^3$, $x^3$ and $y^3$ should be divisible by $7$. But in this case, $xy(x+y)$ is divisible by $7$ and this is a contradiction. So, the only possibility is to have $n=0$ and consequently, $xy(x+y)+4=2$, or, equivalently, $$xy(x+y)+4=-2.\text{\hspace{1em}(7)}$$ The solutions for this are $(x,y)\in \{(-1,-1),(2,-1),(-1,2)\}$, so the required triples are $(a,b,c)=(k+2,k+1,k)$, $k\in \mathbb{Z}$, and all their cyclic permutations.

Alternative version: If $n>0$ then $9$ divides $(a-b)(b-c)(c-a)+4$, that is, the equation $$xy(x+y)+4\equiv 0(\mathrm{mod}9)$$ has the solution $x=b-a$, $y=c-b$. But then $x$ and $y$ have to be $1$ modulo $3$, implying $xy(x+y)\equiv 2(\mathrm{mod}9)$, which is a contradiction. We can continue now as in the first version.