The following problem is taken from the first Benelux Mathematical Olympiad which occurred in 2009.
Let $n$ be a positive integer and let $k$ be an odd positive integer. Moreover, let $a$, $b$ and $c$ be integers (not necessarily positive) satisfying the equation $$a^n+kb=b^n+kc=c^n+ka.$$ Prove that $a=b=c$.
I tried to analyze some congruences module $k$, $a$, $b$ and $c$, but it seems that these relations will not help sufficiently. Also, I did not found a solution for this problem. You can access all the other problems from other years at the BxMo site.
It is clear from the equations that if two among $a, b, c$ are equal, all three must be.
So suppose they are all distinct. Then we have from the equations $$k = \frac{b^n-a^n}{b-c}= \frac{c^n-b^n}{c-a} = \frac{a^n-c^n}{a-b} \tag{1}$$
Among the three $a, b, c$, we must have two of the same parity. WLOG let $a \equiv b \pmod 2$. Then for $k$ to be an odd integer, from $(1)$, we must have $c$ also of the same parity.
Similarly, now among $a \equiv b \equiv c \pmod 2$, we must have two which are equivalent $\pmod 4$. Again from $(1)$, this would force the third to also be equivalent $\pmod 4$.
Continuing in this fashion, we can have $a \equiv b \equiv c \pmod {2^m}$, for some integer $2^m > \max(|a|, |b|, |c|)$, say, which is absurd.