# proving : $a_{1}=a_{2}=\cdots=a_{n}$ when satisfied relation

Suppose $k_1,k_2,k_3,\ldots,k_n$ are non-negative integer numbers such that sum $k_1+k_2+\cdots+k_n$ is an odd number. Let $a_1,a_2,\ldots,a_n$ be arbitrary numbers satisfied: $$\frac {|a_1-a_2|}{k_1}=\frac {|a_2-a_3|}{k_2}=\cdots=\frac {|a_{n-1}-a_n|}{k_{n-1}}=\frac {|a_n-a_1|}{k_n}$$ How Prove that: $a_1=a_2=\cdots=a_n$?

• Arbitrary numbers you say? – anon Jul 8 '12 at 15:55

## 2 Answers

Let us assume that arbitrary numbers means arbitrary real numbers and that the conclusion is false. Then, all the numbers $a_i$ are distinct hence there exists some signs $s_i=\pm1$ and a nonzero real number $c$ such that $s_i\cdot k_i=c\cdot(a_i-a_{i+1})$ for every $1\leqslant i\leqslant n$, with the convention that $a_{n+1}=a_1$. Summing these relations, one gets $\sum\limits_{i=1}^ns_ik_i=0$. For every $i$, $k_i=-k_i\pmod{2}$, hence this implies that $\sum\limits_{i=1}^nk_i=0\pmod{2}$, which contradicts the assumption that $\sum\limits_{i=1}^nk_i$ is odd.

• How did you conclude that $k_{i}=-k_{i}\pmod{2}$? – Frank Jul 8 '12 at 19:30
• For every integer $j$, $j=-j\pmod{2}$ because $j+j=2j=0\pmod{2}$. – Did Jul 8 '12 at 19:33

Edit: You must specify that $a_i\in \mathbb{R}$, since it is false in $\mathbb{C}$.

The statement is false if we are allowed to take $a_i\in \mathbb{C}$. Let $a_0=1$, $a_1=e^{2\pi i/3}$, and $a_2=e^{4\pi i/3}=a_1^2$ so that we are looking at cube roots of unity. Since they form an equilateral triangle in the complex plane, it is clear that $$|a_1-a_0|=|a_2-a_1|=|a_0-a_2|.$$ Thus we are taking each $k_i=1$, so that $k=3$. But this is an odd number. The same reasoning applies for any $n$-gon with the $n^{th}$ roots of unity.