I have a question.

I have been given this proof: "For any $n$ in the integers where $n>2$, show there are at least $2$ elements in $U(n)$ that satisfy $x^2=1$."

I have gone through and actually proved this, (that the numbers are $1$ and $n-1$) but i didn't' know how to prove that $n-1$ is in fact in the set $U(n)$. Is it because two consecutive numbers are always relatively prime?

  • 3
    $\begingroup$ What is $U(n)$? $\endgroup$ – Sigur Feb 7 '13 at 0:22
  • 2
    $\begingroup$ @Sigur: $U(n)$ is the multiplicative group of integers mod $n$ that are relatively prime to $n$. $\endgroup$ – Brian M. Scott Feb 7 '13 at 0:23
  • $\begingroup$ Yes, that is why, and Jason Bourne has just supplied a proof. $\endgroup$ – Brian M. Scott Feb 7 '13 at 0:24
  • $\begingroup$ $-1$ is an integer. $\endgroup$ – jspecter Feb 7 '13 at 0:31

$n$ is coprime to $n-1$, for if $d$ divides $n$ and $d$ divides $n-1$, then $d$ divides $n-(n-1)=1$.

  • $\begingroup$ ah ! such a nice, neat, simple proof. thank you so much $\endgroup$ – Sam Feb 7 '13 at 0:24
  • $\begingroup$ Nice, and concise! (I'd say well-done, and concise, but nice rhymes with concise!) +1 $\endgroup$ – Namaste Feb 7 '13 at 1:11
  • $\begingroup$ $n$ is coprime to $n−2$, for if $d$ divides $n$ and $d$ divides $n−2$, then $d$ divides $n−(n−2)=2$ ? $\endgroup$ – alancalvitti Feb 7 '13 at 2:14
  • $\begingroup$ Not coprime obviously but I suppose if d divides n and n-2 then d=2 thus n is even. $\endgroup$ – ktbiz Feb 18 '16 at 3:24

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