nth roots of unity — A Romanian mathematics olympiad problem

Ref: Titu Andreescu

Problem:

Let $U_{n}$ be the set of nth roots of unity. Prove that the following statements are equivalent:

a) there is $\alpha \in U_{n}$ such that $1+ \alpha \in U_{n}$.

b) there is $\beta \in U_{n}$ such that $1 - \beta \in U_{n}$.

(Romanian Mathematical Olympiad — Second Round, 1990).

Solution:

Assume that there exists $\alpha \in U_{n}$ such that $1 + \alpha \in U_{n}$. Setting $\beta = \frac{1}{1+\alpha}$ we have $\beta^{n} = (\frac{1}{1+\alpha})^{n} = \frac{1}{(1+\alpha)^{n}}=1$, hence, $\beta \in U_{n}$. On the other hand, $1-\beta = \frac{\alpha}{1+\alpha}$ and $(1-\beta)^{n}=\frac{\alpha^{n}}{(1+\alpha)^{n}}=1$, hence, $1- \beta \in U_{n}$, as desired.

Conversely, if $\beta$, $1- \beta \in U_{n}$, set $\alpha = \frac{1-\beta}{\beta}$. Since $\alpha^{n}= \frac{(1-\beta)^{n}}{\beta^{n}}=1$, and $(1-\alpha)^{n}=\frac{1}{\beta^{n}}=1$, we have $\alpha \in U_{n}$, and $1 + \alpha \in U_{n}$, as desired.

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More later,

Nalin Pithwa

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