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I'm asked to find the multiplicative order of $a=860$ in $\langle\mathbb{Z}^*_n, \cdot_n, 1\rangle$, where $n=1383$. Knowing that $ord(a)$ has to be a divisor of $\varphi(n)$, I calculate this number along with its prime factors to get $2^3\cdot5\cdot23=920$. And this is about as far as I get on my own.

In the solution it states, without further explanation, that $ord(a)=230$, which, as far as I can see, is just the least common multiple of $\varphi(n)$'s prime factors. What am I missing?

Omar Sharaki
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  • You can compute $a^{5\cdot 23}\equiv -1 \bmod n$, so that the order of $a$ is $2\cdot5\cdot 23$. – Dietrich Burde Mar 17 '19 at 20:04
  • Could you elaborate your thought process further? – Omar Sharaki Mar 22 '19 at 09:44
  • You only need to know how to calculate $800^{5\cdot 23}$ modulo $1383$. For this, please have a look at this site, e.g., at this duplicate. – Dietrich Burde Mar 22 '19 at 09:47
  • Thank you. Not knowing beforehand, however, that the answer is 230, is $5\cdot23$ an obvious choice? – Omar Sharaki Mar 22 '19 at 11:37
  • Yes, the "larger" divisors of $\phi(1383)=2^3\cdot 5\cdot 23$ are an obvious choice, and there aren't so many. – Dietrich Burde Mar 22 '19 at 12:07
  • How can I be sure though that I haven't overshot $ord(a)$, i.e. how do I know there isn't a smaller product, say k, out of $2^3, 5, 23$ that also leads to $a^k\equiv1$? – Omar Sharaki Mar 22 '19 at 12:24
  • A smaller exponent $k$ with $a^k\equiv 1$ necessarily would have to divide $230$. Now just try the next smallest divisor of $230$. As I said, there aren't too many divisors, really. If a smaller divisor had $a^k=1$, then also $a^{mk}=(a^k)^m=1$. So if you know that $a^{r}=-1$, then $a^k=1$ for $0<k<r$ is impossible. – Dietrich Burde Mar 22 '19 at 12:32

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