For the IMO shortlist 2008, a similar problem (N1) was proposed but $k$ was a prime (this problem is from Benelux 2009 Olympiad, held on May 2009). User Johan Gunardi posted a proof of this already here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1673881#p1673881

WakeUp wrote: 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 equations \[a^n+kb=b^n+kc=c^n+ka \] Prove that $a=b=c$. I have been a bit confused Is there any fault? Let wlog $a\ge b\ge c$ Then $a^n-b^n=k(c-b)\implies c\ge b\implies b=c$ Likewise $a=b=c$ Is this so easy indeed?

You can't w.l.o.g. assume that $a \geq b \geq c$ because it's not symmetric. You can only assume that $a$ is maximal.

ZetaX wrote: You can't w.l.o.g. assume that $a \geq b \geq c$ because it's not symmetric. You can only assume that $a$ is maximal. WakeUp wrote: 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 equations \[a^n+kb=b^n+kc=c^n+ka \] Prove that $a=b=c$. Let $a=max(a,b,c)$ $a^n-b^n=k(c-b)\implies c\ge b$ $0\le c^n-b^n=k(c-a)\le 0$ So $c=b$ then $a=b=c$

You can not conclude $a^n\ge b^n$ from $a\ge b$ since $n$ may be even and $a$ or $b$ may be negative.

1. $ n $ is odd. WLOG $ b \geq c $. then $ a \leq b $ and $ c \geq a $. now we also have $ b \leq c $ so $ a=b=c $. 2. $ n=2m $ $ a^{2m}-b^{2m}=k(c-b) $ $ b^{2m}-c^{2m}=k(a-c) $ $ c^{2m}-a^{2m}=k(b-a) $. it is easy to see that $ a,b,c $ are congruent modulo $ 2 $. let be $ v \geq 1 $ the maximum integer such that $ 2^{v} $ divides $ a-b $. then $ 2^{v+1} $ divides $ b-c $, $ 2^{v+2} $ divides $ c-a $, $ 2^{v+3} $ divides $ b-a $ so $ a-b $ is divisible by every power of $ 2 $ so $ a=b $. in a similar way we obtain that $ b=c $ so $ a=b=c $.

my solution: Note that $\frac{a^n-b^n}{a-b}.\frac{c^n-a^n}{c-a}.\frac{b^n-c^n}{b-c}=-k^3$ so one of $a,b,c$ must be negative. More over $n$ can't be odd because otherwise each of the fractions would be positive which is absurd. Because $2\nmid \frac{a^n-b^n}{a-b}=a^{n-1}+a^{n-2}b+\cdots +b^{n-1}$ numbers $a,b$ must be in different parity or in other words $2\nmid a-b$ similarly we deduce that $2\nmid b-c,2\nmid c-a$ but this is impossible because among any three integers two of them are in the same parity. So $a=b=c$. DONE