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Let $R$ be the set of primitive $42^{\text{nd}}$ roots of unity, and let $S$ be the set of primitive $70^{\text{th}}$ roots of unity. How many elements do $R$ and $S$ have in common?

How would you work this out?

Thanks

Rohini Sharma
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  • $n/42=m/70\Longrightarrow n/m=42/70=3/5$ and $n=3k$; $0\le n < 42$; and we have 14 elements – Michael Galuza Jun 29 '15 at 18:47
  • This question/post would not be a duplicate since this is the problem relating to primitive roots of unity. The other post linked relates to the non-primitive roots of unity. Instead, this should be linked: http://math.stackexchange.com/questions/648822/complex-numbers-and-primitive-roots-of-unity. – Math is Life Jul 10 '15 at 02:52

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Any primitive $42$-th root of unity has order $42$. Any primitive $70$-th root of unity has order $70$. So the sets $R$ and $S$ are disjoint.

We get something more interesting if we ask how many complex numbers are simultaneously a $42$-th root of unity and a $70$-th root of unity (there are $\gcd(42,70)$ of them).

André Nicolas
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