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I'm stuck on proving a rather elementary property, as I'm not really sure how to start off the approach. Suppose $g^a\equiv 1$ mod $m$ and $g^b\equiv 1$ mod $m$. Does this imply that $g^{\gcd(a,b)}\equiv 1$ mod $m$?

Here's my attempt: By definition, we know that $m\mid g^a-1$ and $m|g^b-1$, so there exists some $x,y\in\mathbb{Z}$ such that $mx=g^a-1$ and $my=g^b-1$. Then \begin{align} m(x+y)&=g^{\gcd(a,b)}g^{\frac{a}{\gcd(a,b)}}+g^{\gcd(a,b)}g^{\frac{b}{\gcd(a,b)}}-2\\ &=g^{\gcd(a,b)}\Big(g^{\frac{a}{\gcd(a,b)}}+g^{\frac{b}{\gcd(a,b)}}\Big)-2. \end{align} However, I feel like this approach is only making the problem more complicated than it actually is, as the terms become harder and harder to manipulate to get our desired result $mz=g^{\gcd(a,b)}-1$ for some $z\in\mathbb{Z}.$

Any help would be appreciated!

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    $\begingroup$ By Bezout's, there exists $u,v$ such that $au+bv=\gcd(a,b)$. Thus mod $m$: $g^{\gcd(a,b)}\equiv g^{au+bv}\equiv\cdots$ $\endgroup$
    – anon
    Commented Jun 21, 2012 at 6:48
  • $\begingroup$ Thanks. As described, I just wasn't sure which of the many properties to try out, but this one proves it quite easily. $\endgroup$ Commented Jun 21, 2012 at 7:14
  • $\begingroup$ When you have several things to try, you should actually try them rather than become paralyzed with indecision. You really shouldn't be stuck on a problem until you're at the point where you can't think of anything reasonable to try. $\endgroup$
    – user14972
    Commented Jun 21, 2012 at 9:01
  • $\begingroup$ Also, if it's choosing among many options that's giving you trouble, then you should say so! Without prompting, very few people will even think about offering advice on that topic. In this case, I suspect the answer most people would give is "there was one or more obvious things to try, so I tried them and it was quickly clear that <method of choice> would work". Of course, what's obvious to them might not have been obvious to you -- so such advice may not be helpful if you don't share what alternatives you saw and what you thought about them. $\endgroup$
    – user14972
    Commented Jun 21, 2012 at 9:06
  • $\begingroup$ Sure, I understand, but this comes at the cost of being long-winded, making it harder for others to contribute to the problem at hand without having understood (or at least glimpsed through) the majority of what I would have written down. I feel the current content was accurate enough to explain my overally situation without becoming verbose. $\endgroup$ Commented Jun 21, 2012 at 9:35

3 Answers 3

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We have by the extended Euclidean algorithm that there exists $x, y \in \Bbb{Z}$ such that $ax + by = gcd(a,b)$. So $$ g^{gcd(a,b)} = g^{ax+by} = g^{ax} \cdot g^{by} = (g^a)^x \cdot (g^b)^y \equiv 1^x\cdot 1^y = 1 \pmod{m}. $$

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  • $\begingroup$ Ah, thank you. There's many different properties on $\gcd$ and I wasn't necessarily sure which one would provide the simplest proof. This makes perfect sense though, since we can easily split them in the manner you wrote out. $\endgroup$ Commented Jun 21, 2012 at 7:13
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We do not even need Bézout's Theorem. All we need is to know that the original Euclidean Algorithm (with subtraction) terminates with the $\gcd$. Let $a\gt b$, and let $c=a-b$. Note that $\gcd(b,c)=\gcd(a,b)$.

We have $g^{c}g^{b}=g^{a}$, and therefore if $g^a\equiv 1\pmod{m}$ and $g^b\equiv 1 \pmod{m}$ then $g^c\equiv \pmod{m}$.

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It is easy: $\rm\ \ g^A,\,g^B\equiv 1\:\Rightarrow\,order(g)\, |\, A,B\:\Rightarrow\: order(g)\, |\, (A,B)\:\Rightarrow\: g^{\,(A,B)}\equiv 1\quad$ QED

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