# Complex Roots of Unity and the GCD

I'm looking for a proof of this statement. I just don't know how to approach it. I recognize that $z$ has $a$ and $b$ roots of unity, but I can't seem to figure out what that tells me.

If $z \in \mathbb{C}$ satisfies $z^a = 1$ and $z^b = 1$ then $z^{gcd(a,b)} = 1$.

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If $z^a = 1$ and $z^b = 1$, then also $z^{a+b} = 1$, or more generally $z^{ax+by} = 1$ for all $x, y \in \mathbb{Z}$. – Qiaochu Yuan Mar 1 '12 at 21:55
Unfortunately, this was all the information given in the statement I seek to prove. – Dominick Gerard Mar 1 '12 at 21:56
Do you know that if the gcd is $d$, there are integers $x$ and $y$ such that $ax+by=d$? Then $z^d=z^{ax+by}$. Continue. – André Nicolas Mar 1 '12 at 21:57
Is this logic sound? If $z^a = 1$, then $z^{ax} = 1$. Likewise if $z^b = 1$, then $z^{by} = 1$. Multiplying them together, we get $z^{ax+by} = 1$. By Bezout's we know $ax+by = gcd(a,b)$, so it must be true that $z^{gcd(a,b)} = 1$. – Dominick Gerard Mar 1 '12 at 22:01
@DominickGerard: Yes, those are the steps you should write down; but there are a couple important points to improve the presentation of your proof. You should explain what $x$ and $y$ are when you introduce them. Notice that they have to be integers-- it wouldn't necessarily work with $x=\frac{1}{3}$, for example. In particular, you want them to be integer solutions to $ax+by=\rm{gcd}(a,b)$. This equation certainly isn't true for just any integers $x$ and $y$! I recommend starting off with Bezout's theorem, so that the reader knows just what you mean by $x$ and $y$. – Jonas Kibelbek Mar 1 '12 at 22:27

Hint $\:$ The set of $\rm\:n\in \mathbb Z$ such that $\rm\:z^n = 1\:$ is closed under subtraction so it is closed under $\rm\:gcd$.