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I do not know where to start from this. All I know is a & n are relatively prime so a has an inverse mod n and a belong to Zn. How can I make a connection from this fact to where it said $a^{|u(n)|}$ ? Any thought on this please?

Prove: if gcd(a,n) = 1 then $a^{|u(n)|}$(mod n) = 1 (mod n)

Prove: if $a^{|u(n)|}$(mod n) = 1 (mod n) then gcd(a,n) = 1

Recall: |u(n)| is the order of the group of units modulo n.

This is an if and only if statement so I need to prove both ways. Please help me on how to get start on this?

Amber
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  • Welcome to MSE. In order for your MathJax commands to be effective, they must be surrounded by $ signs. $a^{|u(n)|}$ comes out as $a^{|u(n)|}$ – saulspatz Sep 05 '20 at 02:50
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    You'll find that questions that say only, "Here's my problem, how do I solve it?" don't get a very good response. What do you know about this problem? Any thoughts? – saulspatz Sep 05 '20 at 02:52
  • For the 2nd proof use $,a^k\equiv 1,\Rightarrow, a,$ invertible $\Rightarrow \gcd(a,n)=1,$ by this from the 2nd dupe link. Please search before posting questions. These are FAQs with many prior answers. – Bill Dubuque Sep 05 '20 at 02:52

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