Mathematical Reflections 2014, Issue 2 - Problem J295

Problem:
Let $a,b,c$ be positive integers such that $(a-b)^2+(b-c)^2+(c-a)^2=6abc$. Prove that $a^3+b^3+c^3+1$ is not divisible by $a+b+c+1$.

Proposed by Mihaly Bencze.

Solution:
Clearly $$(a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2-ab-bc-ac)$$ which implies $$a^2+b^2+c^2-ab-bc-ac=3abc.$$
Now, using the identity $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$ we obtain that $$a^3+b^3+c^3-3abc=3abc(a+b+c),$$ or $$a^3+b^3+c^3=3abc(a+b+c+1)$$ which means that $a+b+c+1$ divides $a^3+b^3+c^3$. If $a+b+c+1$ divides $a^3+b^3+c^3+1$, then $a+b+c+1$ divides $(a^3+b^3+c^3+1)-(a^3+b^3+c^3)=1$, which is impossible, since $a+b+c+1>1$. Thus, $a^3+b^3+c^3+1$ is not divisible by $a+b+c+1$.