Consider a group of relatively prime positive integers that are less than the integer \(n\). The operation is multiplication mod \(n\) Show that there exist at least two integers in the group such that \(x^2=1\). assume that \(n\gt 2\)

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Consider a group of relatively prime positive integers that are less than the integer \(n\). The operation is multiplication mod \(n\) Show that there exist at least two integers in the group such that \(x^2=1\). assume that \(n\gt 2\)

Mathematics
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for \(n=5\), we get the set of relatively prime positive integers \(\{1,2,3,4\}\) \(1^2=1\) and \(4^2=1\) are the two elements that satisfy the given condition
Latex tip :) \[x^2\stackrel{n}{\equiv} 1\]
In group {1,2,.......,n-1} , 1 is identity and 1^2=1 is always in the group. We know that (n-1) is relative prime with n, that is gcd (n-1,n)=1 or \(n-1\equiv 1(mod n)\) \((n-1)(n-1) \equiv 1*1=1(mod n)\) we can use multiplication table of group to show that there is at least 2 elements on the table satisfy the condition. |dw:1439986568531:dw|

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Nice! so you're saying that \(1\) and \(n-1\) are relatively prime to \(n\) and they have order \(2\) no matter what \(n\) is
I think so.
gotcha! thank you :)

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