Find all solutions to $x^2$ ≡ 1 in arithmetic modulo 8.
My understanding is that what this is saying is that find the x which multiplicative inverse is equal to itself within modulo 8.
I know that this is the case for certain examples like in modulo 18: the multiplicative inverse of 17 is also 17. But I am not quite sure why.
Any explanation and help with a technique to solve would be appreciated.
Best Answer
$x^2\equiv 1\ (\mod 8)\\ x^2-1\equiv0\ (\mod 8)\\ (x-1)(x+1)\equiv0\ (\mod 8)\\ 8\mid(x-1)(x+1) $
By trial and error, we get $x\equiv a\ (\mod8)$, where $a\in\{1,7,3,5\}$.