Consider the sequence of numbers
$ \ a,a^2, a^4, a^8 , ............., \ (mod \ N) \ $
such that $ gcd(a,N)=1 \ $ and $ \ N \ $ is an odd prime. Then
(i) Prove that $ \exists \ $ two terms of the sequence having the same value.
(ii) Prove that $ \ a^x \equiv 1 \ (mod \ N) \ $
Answer:
(i)
Given $ \ gcd(a,N)=1 \ $
Then there exits $ \ m,n \in \mathbb{Z} \ $ such that $ \ ma+nN=1 \ $
We will show that for any $ n=2l , \ l \in \mathbb{N} , \ \ \ a^{2l}=(a^{2l})^{2} \ \ (mod \ N) $
Since $ \ N \ $ is odd prime , we have
$$ N \neq 2 $$
But next I can not prove it.
Help me doing this.
$N=3$ is an odd prime, $a=2$ is coprime with $N$, yet $a\equiv2\bmod3$ and $a^2\equiv1\bmod3$.
What you probably mean is not that any two terms have the same value, but there are two terms which have the same value. This is a consequence of the pigeonhole principle, given that the sequence is infinite and there are finitely many residues modulo $N$.
As for the second question, what you probably mean is that this equation can be solved for $x$. To do so, from Part 1. you have $a^p\equiv a^q$, assuming $p<q$, you can deduce $1\equiv a^{q-p}\mod N$.