So some of you may have noticed my algebra exam is tomorrow!
The question is: If $[a]_{n} = [1]_{n}$ prove that $\mathrm{gcd}(a, n) = 1$.
Well I tried it one way. We know the above means the following: $a \equiv 1 \pmod{n}$. This is also written as $1 \equiv a \pmod{n}$ Then the gcd of these are as follows: $\mathrm{gcd}(a, n) := d $ and $\mathrm{gcd}(1, n) := d'.$
At this point I thought $ d = d'$ which is NOT the case as pointed out by roommate. but lets assume it does so I finished the proof as follows:
$\mathrm{gcd}(a,n) = \mathrm{gcd}(1, n)$ $\mathrm{gcd}(a, n) = 1$
but then obviously that's not the case (or is it). I then tried the following: $d = as + nt, \quad s, t \in \mathbb{Z}$ $d' = as' + nt', \quad s', t' \in \mathbb{Z}.$
And now I'm stuck again :(