How to get common roots of unity of $ z^{6}=1 $ and $ z^{21}=1 $?

Question

How to get common roots of unity of $z^{21}=1$ and $z^{6}=1 $ for $ z\in\mathbb C $?

I know that $ z^{n} =1 $ has roots $ z=e^{\frac{2\pi k }{n}i} $ where $ k\in \{0,1,2,...,n-1\} $

Answer 1

$$ \gcd( 21, 6 ) = ??? $$

$$ \frac{ 21 }{ 6 } = 3 + \frac{ 3 }{ 6 } $$ $$ \frac{ 6 }{ 3 } = 2 + \frac{ 0 }{ 3 } $$ Simple continued fraction tableau:
$$ \begin{array}{cccccc} & & 3 & & 2 & \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 3 }{ 1 } & & \frac{ 7 }{ 2 } \end{array} $$ $$ $$ $$ 7 \cdot 1 - 2 \cdot 3 = 1 $$

$$ \gcd( 21, 6 ) = 3 $$
$$ 21 \cdot 1 - 6 \cdot 3 = 3 $$

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a different example:

$$ \gcd( 7654321, 1234567 ) = ??? $$

$$ \frac{ 7654321 }{ 1234567 } = 6 + \frac{ 246919 }{ 1234567 } $$ $$ \frac{ 1234567 }{ 246919 } = 4 + \frac{ 246891 }{ 246919 } $$ $$ \frac{ 246919 }{ 246891 } = 1 + \frac{ 28 }{ 246891 } $$ $$ \frac{ 246891 }{ 28 } = 8817 + \frac{ 15 }{ 28 } $$ $$ \frac{ 28 }{ 15 } = 1 + \frac{ 13 }{ 15 } $$ $$ \frac{ 15 }{ 13 } = 1 + \frac{ 2 }{ 13 } $$ $$ \frac{ 13 }{ 2 } = 6 + \frac{ 1 }{ 2 } $$ $$ \frac{ 2 }{ 1 } = 2 + \frac{ 0 }{ 1 } $$ Simple continued fraction tableau:
$$ \begin{array}{cccccccccccccccccc} & & 6 & & 4 & & 1 & & 8817 & & 1 & & 1 & & 6 & & 2 & \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 6 }{ 1 } & & \frac{ 25 }{ 4 } & & \frac{ 31 }{ 5 } & & \frac{ 273352 }{ 44089 } & & \frac{ 273383 }{ 44094 } & & \frac{ 546735 }{ 88183 } & & \frac{ 3553793 }{ 573192 } & & \frac{ 7654321 }{ 1234567 } \end{array} $$ $$ $$ $$ 7654321 \cdot 573192 - 1234567 \cdot 3553793 = 1 $$

Answer 2

$$ \begin{align}\{ z : z^6=z^{21}=1, z\in\mathbb C \}&\Longleftrightarrow \{ k : k=o(z),\ k|6,\ k|21 \}\\ &\Longleftrightarrow\{ k : k=o(z),\ k|\gcd (6, 21) \}\\ &\Longleftrightarrow\{ k : k=o(z),\ k|3 \}\\ &\Longleftrightarrow\{ k : k=o(z),\ k=1\ \text{or}\ k=3 \}\\ &\Longleftrightarrow\{ z\in \mathbb C : z^3=1 \}\\ &\Longleftrightarrow\{ z : z=1\ \text{or}\ z=e^{\frac{2\pi i}{3}}\ \text{or}\ z=e^{\frac{4\pi i}{3}} \} \end{align}$$

Answer 3

$z$ will satisfy both of $z^{21}=1$ and $z^6=1$ iff $z^3=1$. One direction is trivial, the other follows from $z^6=1\implies z^{21}=z^3$.

Thus $ z\in \{ 1,e^{\frac{2\pi i}3},e^{\frac{4\pi i}3} \} $.