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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.

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1 Answer 1

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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.

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  • $\begingroup$ Shouldn't it be $\equiv$, not $=$? $\endgroup$
    – S.C.B.
    Commented Mar 4, 2016 at 6:28
  • $\begingroup$ Also, shouldn't it be ''we can have $a\equiv b\equiv c\;\text{mod}\;2^m$'' instead of ''we can have $2^m$ be a factor for all three''? $\endgroup$
    – Larara
    Commented Mar 4, 2016 at 6:42
  • $\begingroup$ @Larara oh yes. Corrected. $\endgroup$
    – Macavity
    Commented Mar 4, 2016 at 6:43
  • $\begingroup$ @Macavity one thing that I noticed is that we have to use the fact that if $k$ is odd, then $\bar{k}$ (where the bar denotes the residue class modulo $2^n$) doesn't have any divisors of $0$ that are even in $\frac{\mathbb{Z}}{2^n\mathbb{Z}}$, i.e., if $\bar{s}\bar{k}\equiv 0\;\text{mod}\,2^n$, with $s\equiv 0\;\text{mod}\,2$, then $s\equiv 0\;\text{mod}\, 2^n$. We have to use this to conclude that $a$, $b$ and $c$ are equivalent $\text{mod}\,2^n$ for any $n\geq 1$. $\endgroup$
    – Larara
    Commented Mar 5, 2016 at 3:35
  • $\begingroup$ At each step, all we need to use is the Pigeonhole principle. $\endgroup$
    – Macavity
    Commented Mar 5, 2016 at 4:01

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