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.

*******************************************************************************************************

More later,

Nalin Pithwa

 

 

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s