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Suppose $ord_{341} 171 = t$. Show that for any positive integer k, $ord_{341} (171^{k}) = t $ if and only if k is an odd integer that is not a multiple of 5.

This doesnt seem right useing the CRT it implies that $ord_{341} 171 $ we have that $ 171^{10} \equiv 1 \mod11 $ and $ 171^{30} \equiv 1 \mod31 $ which implys that $ 171^{30} \equiv 1 \mod341 $ which is clearly a multiple of 5... in fact i thought i found the least such interger was $10$ which is supposed to be the order no; what am i doing wrong?

Faust
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You aren't doing anything wrong: $\text{ord}_{341}(171)=10$. Now you need to show that $\text{ord}_{341}(171^k)=10$ exactly when $k$ is an odd integer that is not a multiple of $5$. One nice way to do that is suppose $k=10r+s$, with $r$ a positive integer, and $0 \le s <10$. You can quickly show that you have but $10$ cases to consider.

Matthew Conroy
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    Weird i was not understanding the question, is there a specific name for the set of integers that solve this? – Faust Oct 27 '17 at 20:23
  • Perhaps, but I don't recall any. My elementary group theory nomenclature memory has not been refreshed in some time! Cheers! – Matthew Conroy Oct 27 '17 at 20:30