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ganeshie8
 one year ago
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 \(n\gt 2\)
ganeshie8
 one year ago
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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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2for \(n=5\), 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Latex tip :) \[x^2\stackrel{n}{\equiv} 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In group {1,2,.......,n1} , 1 is identity and 1^2=1 is always in the group. We know that (n1) is relative prime with n, that is gcd (n1,n)=1 or \(n1\equiv 1(mod n)\) \((n1)(n1) \equiv 1*1=1(mod n)\) 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Nice! so you're saying that \(1\) and \(n1\) are relatively prime to \(n\) and they have order \(2\) no matter what \(n\) is

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2gotcha! thank you :)
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