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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 \(n\gt 2\)

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  1. ganeshie8
    • one year ago
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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

  2. anonymous
    • one year ago
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    Latex tip :) \[x^2\stackrel{n}{\equiv} 1\]

  3. anonymous
    • one year ago
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    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|

  4. ganeshie8
    • one year ago
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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

  5. anonymous
    • one year ago
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    I think so.

  6. ganeshie8
    • one year ago
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    gotcha! thank you :)

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